CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Nernst equation Question Bank Set 1 | CHEM 107 - Essentials of General and Organic Chemistry

CHEM 107 - ESSENTIALS OF

GENERAL AND ORGANIC

CHEMISTRY - Nernst equation

Question Bank - Set 1

Liberty University

Question 1

Question

Calculate the cell potential of a galvanic cell at 25

C where the concentrations

of Mn2+ and MnO−

4are 0.10 M and 1.0×10−5M, respectively. The standard

reduction potential for the half-reaction MnO−

4+ 8H++ 5e−→Mn2+ + 4H2O

is 1.51 V. (Assume that the temperature is 25

C and the cell is at standard

conditions.)

Solution

Step 1: Write the balanced redox reaction for the cell and determine the cell

potential under standard conditions.

MnO−

4+ 8H++ 5e−→Mn2+ + 4H2O

The standard cell potential (E◦) under standard conditions can be calculated

using the Nernst equation:

E◦=E◦

cathode −E◦

anode

Given that E◦= 1.51 V and the cell is galvanic (spontaneous), E◦>0.

Step 2: Calculate the cell potential under non-standard conditions using the

Nernst equation. The Nernst equation is given by:

E=E◦−0.0592

nlog [Mn2+]

[MnO−

4]5[H+]8

Substitute the given values into the equation:

E= 1.51 V −0.0592

5log 0.10

(1.0×10−5)5(1)8

E= 1.51 V −0.0592

5log 0.10

1.0×10−25 

E= 1.51 V −0.0592

5log1025

E= 1.51 V −0.0592

5×25

E= 1.51 V −0.1184

E= 1.3916 V

Therefore, the cell potential of the galvanic cell at 25

C is 1.3916 V.

Question 2

Question

Calculate the cell potential for the following redox reaction at 25

Zn(s) + F e2+(aq)→Zn2+(aq) + F e(s)

given that [F e2+] = 0.10 M, [Zn2+]=1.0 M, and the standard reduction poten-

tials are E0

Fe2+/Fe =−0.44 V and E0

Zn2+/Zn =−0.76 V.

Solution

Step 1: Write the half-reactions and the overall cell reaction. The half-reactions

are:

Cathode (Reduction): F e2+(aq)+2e−→F e(s)E0

cathode =−0.44 V

Anode (Oxidation): Zn(s)→Zn2+(aq)+2e−E0

anode =−0.76 V

The overall reaction is the sum of the two half-reactions:

Zn(s) + F e2+(aq)→Zn2+(aq) + F e(s)

Step 2: Calculate the cell potential using the Nernst equation: The Nernst

equation is given by:

Ecell =E0

cell −0.0592

nlog Q

where: - Ecell is the cell potential, - E0

cell is the standard cell potential, - nis the

number of electrons transferred in the balanced cell reaction, - Qis the reaction

quotient, and - Kis the equilibrium constant.

For this reaction, n= 2 as 2 electrons are transferred. The reaction quotient

Qfor this reaction is:

Q=[Zn2+]

[F e2+]=1.0

0.10 = 10

Substitute the values into the Nernst equation:

Ecell = (−0.76 V) −0.0592

2log(10)

Ecell =−0.76 V −0.0296 log(10)

Ecell =−0.76 V −0.0296 ×1

Ecell =−0.76 V −0.0296

Ecell =−0.7896 V

Therefore, the cell potential for the given redox reaction at 25

C is −0.7896 V.

Question 3

Question

A galvanic cell contains a copper electrode in a 1.0 M solution of Cu2+ ions and

a silver electrode in a 0.10 M solution of Ag+ions. Calculate the cell potential

at 25

C given that E◦

cell = 0.46 V.

Solution

Step 1: Write the half-reactions for the cathode and anode. The reduction half-

reactions for the copper electrode (Cu2+ gaining electrons) and the silver elec-

trode (Ag+gaining electrons) are: Cathode: Cu2+(aq)+2e−→Cu(s)E◦

cathode =

0.34 V Anode: Ag+(aq) + e−→Ag(s)E◦

anode = 0.80 V

Step 2: Calculate the cell potential using the Nernst equation. The Nernst

equation is given by:

Ecell =E◦

cell −0.0592

nlog [reduced form (cathode)]m

[oxidized form (anode)]n

where: n= number of electrons transferred (n= 1 in this case) m= stoichio-

metric coefficient of the reduced form (cathode)

Ecell =E◦

cell −0.0592

1log [Cu(s)]

[Ag(s)] 

Step 3: Calculate the concentration of Cu2+ and Ag+. Given: [Cu2+] =

1.0 M and [Ag+] = 0.10 M

Step 4: Substitute the concentrations into the Nernst equation and solve for

Ecell.

Ecell = 0.46 V −0.0592

1log 1

0.10

Ecell = 0.46 V −0.0592 ×log(10)

Ecell = 0.46 V −0.0592 ×1

Ecell = 0.46 V −0.0592 V = 0.40 V

Therefore, the cell potential at 25

C is 0.40 V.

Question 4

Question

The concentration of Br−ions in a solution is 1.5×10−2M. The standard

reduction potential of the Br2/Br−half-reaction is +1.07 V. Calculate the cell

potential when the concentration of Br−ions is diluted to 5.0×10−3M at 25

(Assume the temperature coefficient αis 0.0 V/

C.)

Solution

Step 1: Write the Nernst equation. The Nernst equation relates the cell potential

to the concentrations of the reactants and products in an electrochemical cell:

E=E◦−0.0592

nlog [Reductant]

[Oxidant]

where: - Eis the cell potential, - E◦is the standard reduction potential, - nis

the number of electrons transferred in the half-reaction, and - [Reductant] and

[Oxidant] are the concentrations of the reductant and oxidant, respectively.

Step 2: Determine the number of electrons transferred, n. Since the half-

reaction involves the reaction Br2+2e−→2Br−, the number of electrons trans-

ferred is 2.

Step 3: Calculate the cell potential at the new concentration. First, calculate

the new cell potential using the Nernst equation:

E= 1.07 V −0.0592

2log 1.5×10−2

5.0×10−3

Step 4: Substitute the values and solve for E.

E= 1.07 V −0.0592

2log 1.5×10−2

5.0×10−3= 1.07 V −0.0592

2log 3

E= 1.07 V −0.0296 log 3

Step 5: Calculate the final cell potential.

E= 1.07 V −0.0296 ×0.4771 = 1.07 V −0.0141 = 1.0559 V

Therefore, the cell potential when the concentration of Br−ions is diluted

to 5.0×10−3M at 25

C is 1.0559 V.

Question 5

Question

A concentration cell is set up with two half-cells. One half-cell has a silver

electrode in a 0.10 M AgNO3solution, and the other half-cell has a silver

electrode in a 0.0010 M AgNO3solution. Calculate the cell potential for this

concentration cell at 298 K. (Given: E◦

cell = 0.80 V)

Solution

Step 1: Write the half-reactions involved in the concentration cell. The two half-

reactions involved are: Anode: Ag(s)→Ag+(aq) + e−(oxidation) Cathode:

Ag+(aq) + e−→Ag(s) (reduction)

Step 2: Write the Nernst equation for the cell potential (Ecell). The Nernst

equation is given by: Ecell =E◦

cell −0.0592

nlog [Ag+]anode

[Ag+]cathode 

Step 3: Calculate the concentration ratio of [Ag+]anode to [Ag+]cathode.

Given: [Ag+]anode = 0.10 M [Ag+]cathode = 0.0010 M

Concentration ratio: [Ag+]anode

[Ag+]cathode =0.10

0.0010 = 100

Step 4: Substitute the values into the Nernst equation and solve for Ecell.

Ecell = 0.80V−0.0592

1log(100) Ecell = 0.80V−0.0592×2Ecell = 0.80V−0.1184

Ecell = 0.6816 V

Therefore, the cell potential for this concentration cell at 298 K is 0.6816 V.

Question 6

Question

Given the following cell notation representing a galvanic cell:

Mn3+(aq)|Mn2+(aq)|| Cu2+(aq)|Cu(s)

where the reduction potentials are as follows: E0(Mn3+/Mn2+) = −1.18 V

and E0(Cu2+/Cu) = +0.34 V. Determine the cell potential at 25◦Cwhen

[Mn3+]=0.10 M, [Mn2+]=0.20 M, and [Cu2+] = 0.50 M.

Solution

Step 1: Write down the half-cell reactions for the given redox reactions. The

half-cell reactions are:

Cathode: Cu2+(aq)+2e−→Cu(s)

Anode: Mn3+(aq) + e−→Mn2+(aq)

Step 2: Calculate the standard cell potential, E0

cell, using the Nernst equa-

tion:

cell =E0

cathode −E0

anode

cell = (+0.34 V)−(−1.18 V) = 1.52 V

Step 3: Calculate the reaction quotient, Q, using the concentrations given:

Q=[Mn2+]

[Mn3+]×1

[Cu2+]=0.20

0.10 ×1

0.50 = 4.00

Step 4: Calculate the cell potential at 25◦Cusing the Nernst equation:

Ecell =E0

cell −0.0592 V

nlog(Q)

Since the reaction quotient, Q, is 4 and there are 1 electron transferred in this

reaction, we have:

Ecell = 1.52 V−0.0592 V

1log(4)

Ecell = 1.52 V−0.0592 V×0.6021

Ecell = 1.52 V−0.0358 V

Ecell = 1.4842 V

Therefore, the cell potential at 25◦Cis 1.4842 V.

Question 7

Question

Calculate the standard cell potential for the following reaction at 25

Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)

Given:

E◦

Ag+/Ag = 0.80 V

E◦

Zn2+/Zn =−0.76 V

Solution

Step 1: Write the Nernst equation for the cell potential:

E=E◦−RT

nF ln(Q)

Step 2: Calculate the standard cell potential (E◦)

E◦=E◦

cathode −E◦

anode

E◦=E◦

Ag+/Ag −E◦

Zn2+/Zn

E◦= 0.80 V −(−0.76 V)

E◦= 1.56 V

Step 3: Calculate the reaction quotient (Q) for the given cell:

Q=[Zn2+]

[Ag+]2

Step 4: Substitute the values into the Nernst equation and calculate the cell

potential (E):

E= 1.56 V −(8.314J/Kmol)(298K)

2(96485C/mol)ln(Q)

E= 1.56 V −2482.452

1930.7ln(Q)

E= 1.56 V −1.2869 ln(Q)

Therefore, the cell potential for the given reaction at 25

C is given by E=

1.56 V −1.2869 ln(Q).

Question 8

Question

Calculate the cell potential for a galvanic cell in which [Cu2+] = 0.10 M, [Zn2+]

= 1.0 ×10−4M, [Cu] = 1.0 M, and [Zn] = 1.0 M, at 25

C. The standard

reduction potentials are E◦(Cu2+/Cu) = 0.34 V and E◦(Zn2+/Zn) = -0.76 V.

Solution

1. The cell potential at non-standard conditions is calculated using the Nernst

equation:

E = E◦−0.0592

nlog(Q)

where E is the cell potential, E◦is the standard cell potential, n is the

number of moles of electrons transferred, and Q is the reaction quotient.

2. The reaction taking place in the cell is: Zn(s) + Cu2+(aq) →Zn2+(aq) +

Cu(s).

3. The number of moles of electrons transferred in this reaction is 2, because

zinc loses 2 electrons and copper gains 2 electrons.

4. First, calculate the reaction quotient Q:

Q=[Zn2+][Cu]

[Zn][Cu2+]

Q=(1.0×10−4)(1.0)

(1.0)(0.10)

Q= 1.0×10−3

5. Now, substitute the given values into the Nernst equation:

E = 0.34 −0.0592

2log1.0×10−3

E = 0.34 −0.0296 ×3

E=0.34 −0.0888

E=0.2512 V

6. Therefore, the cell potential for the galvanic cell at these non-standard

conditions is 0.2512 V.

Question 9

Question

A concentration cell is constructed using two half-cells. One half-cell has a

silver wire in a silver ion solution with a concentration of 1.0×10−3M, and

the other half-cell has a silver wire in a silver ion solution with a concentration

of 1.0×10−2M. What is the cell potential at 25

C for this concentration cell?

Given that the standard reduction potential for the Ag+—Ag half-cell is 0.80

Solution

Step 1: Write the half-reactions for each half-cell. The two half-reactions in-

volved are:

Ag++e−→Ag (s)

Step 2: Write the Nernst equation for the cell potential. The Nernst equation

is given by:

Ecell =E◦

cell −0.0592

nlog [Ag+]cathode

[Ag+]anode 

where: Ecell = cell potential E◦

cell = standard cell potential n= number of moles

of electrons transferred (1 in this case) [Ag+]cathode = concentration of silver

ion in the cathode half-cell [Ag+]anode = concentration of silver ion in the anode

half-cell

Step 3: Calculate the cell potential. Substitute the given values into the

Nernst equation:

Ecell = 0.80 V −0.0592

1log 1.0×10−3M

1.0×10−2M

Ecell = 0.80 V −0.0592

1log(0.1)

Ecell = 0.80 V −0.0592 log(0.1)

Ecell = 0.80 V −0.0592(−1)

Ecell = 0.80 + 0.0592

Ecell = 0.8592 V

Therefore, the cell potential at 25

C for this concentration cell is 0.8592 V.

Question 10

Question

A cell consists of a zinc electrode in a 2.0 M Zn2+ solution and a copper electrode

in a 0.10 M Cu2+ solution. If the standard reduction potentials are E◦

Zn2+/Zn =

−0.76Vand E◦

Cu2+/Cu = +0.34V, calculate the cell potential at 25

C using the

Nernst equation.

Solution

Let’s use the Nernst equation to calculate the cell potential at 25

E=E◦−0.0592

nlog [Cathode]

[Anode] 

where: E= cell potential E◦= standard cell potential n= number of

electrons transferred [Cathode] = concentration of the cathode species [Anode]

= concentration of the anode species

First, calculate the number of electrons transferred (n) in the reaction. The

half-reactions are:

Zn2+ + 2e−→Zn E◦=−0.76

Cu2+ + 2e−→Cu E◦= +0.34

From the half-reactions, we can see that 2 electrons are transferred in each

reaction.

Step 1: Calculate the cell potential at 25

C for the given concentrations

using the Nernst equation:

E=E◦

cell −0.0592

2log [Cu2+]

[Zn2+]

Step 2: Substitute the given values into the equation:

E= 0.34V−0.0592

2log 0.10

2.0

E= 0.34V−0.0296 log(0.05)

Step 3: Calculate the natural logarithm:

E= 0.34V−0.0296 ×(−2.9957)

E= 0.34V+ 0.0887

Step 4: Calculate the cell potential:

E= 0.43V

Therefore, the cell potential at 25

C using the Nernst equation is 0.43 V.

Question 11

Question

A galvanic cell consists of a silver electrode in contact with a silver ion solution of

unknown concentration, and a zinc electrode in contact with a zinc ion solution

of concentration 0.10 M. The cell potential is measured to be +0.42 Vat 25◦C.

Determine the concentration of silver ions in the silver half-cell.

Solution

Step 1: Write the half-reactions for the cell. The cell consists of the following

two half-reactions: 1. Anode (Zinc electrode): Zn(s)→Zn2+(aq) + 2e−2.

Cathode (Silver electrode): Ag+(aq) + e−→Ag(s)

Step 2: Calculate cell potential using the Nernst equation. The Nernst

equation is given by:

E=E◦−RT

nF ln(Q)

Given that the cell potential Eis +0.42 V, the standard cell potential E◦is

+1.10 Vfor the reduction of Ag+ions to Ag(s). The temperature Tis 25◦C

which is equivalent to 298 K. The number of electrons transferred nin the

reaction is 1, the Faraday constant Fis 96485 C/mol, and the gas constant

Ris 8.314 J/(mol ·K). The reaction quotient Qcan be calculated using the

concentrations of Zn2+ and Ag+ions.

Step 3: Determine the concentration of Ag+ions in the silver half-cell.

To determine the concentration of Ag+ions, we first need to calculate the

value of the reaction quotient Q. Since the reaction is at equilibrium, Q=K

(equilibrium constant) for the cell reaction. The equilibrium constant expression

is given by:

K=[Ag(s)]

[Ag+(aq)]

Since the activity of a pure solid is considered to be 1, the equilibrium con-

stant can be simplified to 1

[Ag+(aq)] . Thus, Q=1

[Ag+(aq)] . Next, substitute the

values into the Nernst equation and solve for [Ag+(aq)]. This will give us the

concentration of silver ions in the silver half-cell.

Question 12

Question

A galvanic cell is constructed with a standard hydrogen electrode (SHE) as the

anode and a silver-silver chloride electrode as the cathode. The concentration

ceAg+ ions in the cathode compartment is 0.10 M and the partial pressure of

ceH2 gas in the anode compartment is 0.50 atm. Calculate the cell potential at

Given: Standard reduction potentials:

ceAg + +e− − > Ag(s)E0= 0.80V

ce2H+ +2e− − > H2(g)E0= 0.00V

Faraday’s constant, F= 96,485 C mol−1, Gas constant, R= 8.314 J K−1mol−1,

Temperature, T= 298 K

Solution

Step 1: Write the half-reactions for the anode and cathode: Anode (oxidation

at SHE):

ceH2(g)−>2H+ +2e−

Cathode (reduction at Ag/AgCl electrode):

ceAg + +e− − > Ag(s)

Step 2: Calculate the cell potential without considering non-standard con-

ditions:

cell =E0

cathode −E0

anode

cell = 0.80 V−0.00 V= 0.80 V

Step 3: Calculate the cell potential under non-standard conditions using the

Nernst equation:

Ecell =E0

cell −RT

nF ln Q

P

Where: Ecell = cell potential under non-standard conditions E0

cell = cell

potential under standard conditions R= gas constant T= temperature n=

number of moles of electrons transferred F= Faraday’s constant Q= reaction

quotient P= atmospheric pressure

Here, n= 2 (moles of electrons transferred)

Ecell = 0.80 V−(8.314 J K−1mol−1)(298 K)

(2)(96485 C mol−1)ln 0.10

0.50

Ecell = 0.80 V−(2476.52)

(2)(96485) ln(0.20) = 0.80 V−(0.0128) ln(0.20)

Ecell ≈0.80 V−0.0128(−0.693) = 0.80 V+ 0.0089 ≈0.81 V

Therefore, the cell potential at 25

C under these non-standard conditions is

approximately 0.81 V.

Question 13

Question

A voltaic cell is set up where the standard reduction potential of the cathode is

+0.42 V and that of the anode is -0.80 V. Calculate the cell potential when the

concentration of Mn2+ is 0.010 M at the anode and the concentration of Ag+

is 2.0 M at the cathode.

Solution

Step 1: Write the half-reactions and their standard reduction potentials. The

half-reactions involved are:

Anode: Mn2+(aq) →Mn(s) + 2e−E0=−0.80 V

Cathode: Ag+(aq) + e−→Ag(s) E0= +0.42 V

Step 2: Calculate the cell potential at standard conditions. The cell potential

at standard conditions is given by the formula:

cell =E0

cathode −E0

anode

cell = (+0.42 V) −(−0.80 V) = 1.22 V

Step 3: Determine the reaction quotient Q. The reaction quotient Q is given

by:

Q=[Mn]

[Ag+]=0.010

2.0= 0.0050

Step 4: Calculate the cell potential at non-standard conditions using the

Nernst equation. The Nernst equation is given by:

Ecell =E0

cell −0.0592

nlog(Q)

where n is the number of electrons transferred in the balanced redox reaction.

Step 5: Calculate the cell potential at non-standard conditions. Since 2

moles of electrons are transferred,

Ecell = 1.22 V −0.0592

2log(0.0050) = 1.22 V −0.0592 ×1×(−2.30) = 1.35 V

Therefore, the cell potential at non-standard conditions is 1.35 V.

Question 14

Question

At 25

C, a cell has a standard cell potential of 1.23 V and an [Cu2+] concentra-

tion of 0.10 M in the cathode compartment while the [Cu2+] concentration in the

anode compartment is 0.001 M. Calculate the cell potential at this nonstandard

condition.

Solution

The Nernst equation is given by:

E=E◦−0.0592

nlog Q

where: - Eis the cell potential at nonstandard conditions (in V) - E◦is the

standard cell potential (in V) - nis the number of moles of electrons transferred

in the balanced cell reaction - Qis the reaction quotient

Step 1: Write the balanced redox reaction for the cell. The cell reaction is:

Cu2+ + 2e−→Cu

Therefore, the number of moles of electrons transferred, n, is 2.

Step 2: Calculate the reaction quotient Q.

Q=[Cu]

[Cu2+]=0.10

0.001 = 100

Step 3: Substitute the known values into the Nernst equation.

E= 1.23 −0.0592

2log 100

Step 4: Simplify the equation.

E= 1.23 −0.0296 log 100 = 1.23 −0.0296 ×2=1.23 −0.0592 = 1.17 V

Therefore, the cell potential at this nonstandard condition is 1.17 V.

Question 15

Question

A concentration cell is set up using two half-cells, both containing standard

hydrogen electrodes. One half-cell has a hydrogen ion concentration of 0.1 M,

while the other half-cell has a hydrogen ion concentration of 0.01 M. Calculate

the cell potential at 25

C using the Nernst equation.

Solution

Step 1: Write the half-reactions and the cell reaction. The half-reactions for the

standard hydrogen electrode are:

Reduction: 2H++ 2e−→H2E◦

red = 0 V

Oxidation: H2→2H++ 2e−E◦

ox = 0 V

The overall cell reaction is the difference between the two half-reactions:

H2(0.1 M) →H2(0.01 M)

Step 2: Calculate the standard cell potential (E◦

cell). Since the standard

reduction potential of the standard hydrogen electrode is 0 V, the standard cell

potential is also 0 V.

Step 3: Calculate the reaction quotient (Q). The reaction quotient, Q, can

be calculated using the concentrations of the hydrogen ions in each half-cell:

Q=[H+

[H+

2]=0.1

0.01 = 10

Step 4: Calculate the cell potential at 25

C using the Nernst equation. The

Nernst equation relates the cell potential (Ecell) to the standard cell potential

(E◦

cell), the gas constant (R), the temperature (T), the number of electrons

transferred (n), and the Faraday constant (F):

Ecell =E◦

cell −RT

nF ln(Q)

Plugging in the values:

Ecell = 0 V −(8.314 J/mol ·K)(298 K)

2(96485 C/mol) ln(10)

Ecell = 0 V −8.314 J/mol ·K·298 K

2·96485 C/mol ln(10)

Ecell = 0 V −6221.252

192970 ln(10)

Ecell ≈ −0.019 V

Therefore, the cell potential at 25

C using the Nernst equation is approxi-

mately -0.019 V.

Question 16

Question

A voltaic cell consists of a standard hydrogen electrode (SHE) and a copper-

silver cell. The concentration of copper ions in the copper half-cell is 0.10 M,

while the concentration of silver ions in the silver half-cell is 1.0 M. Calculate

the cell potential at 25◦C when the copper electrode is the anode. Given:

E◦(Cu2+/Cu) = 0.34 V and E◦(Ag+/Ag) = 0.80 V.

Solution

Step 1: Write the cell reaction and determine the overall cell potential.

The cell reaction is:

Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s)

The overall cell potential (E◦

cell) can be calculated as:

E◦

cell =E◦

cathode −E◦

anode

Given E◦(Cu2+/Cu) = 0.34 V and E◦(Ag+/Ag) = 0.80 V, we have:

E◦

cell =E◦

Ag+/Ag −E◦

Cu2+/Cu

E◦

cell = 0.80 V −0.34 V

E◦

cell = 0.46 V

Therefore, the standard cell potential is 0.46 V.

Step 2: Apply the Nernst equation to calculate the cell potential at non-

standard conditions.

The Nernst equation is given by:

E=E◦−0.0592

nlog Q

Where: - Eis the cell potential under non-standard conditions. - E◦is the

standard cell potential. - nis the number of moles of electrons transferred in

the balanced redox reaction. - Qis the reaction quotient. - Kis the equilibrium

constant.

In this case, n= 2 (from the balanced reaction), K= 1.8×1010, and

Q= [Cu2+]/[Ag+]2.

Step 3: Calculate Q.

Q=0.10

(1.0)2= 0.10

Step 4: Substitute all known values into the Nernst equation and solve for

E= 0.46 V −0.0592

2log 0.10

1.02

E= 0.46 V −0.0296 log 0.10

E= 0.46 V −0.0296 ×(−1)

E= 0.46 V + 0.0296

E= 0.4896 V

Therefore, the cell potential at 25◦C when the copper electrode is the anode

is 0.4896 V.

Question 17

Question

A galvanic cell is set up with a standard hydrogen electrode (SHE) on one side

and a silver electrode on the other side. The standard reduction potential of the

silver electrode is E◦= 0.80 V. If the concentration of Ag+ions is 1.0×10−3M

in the solution, calculate the cell potential at 25◦C using the Nernst equation.

Solution

Step 1: Write down the standard cell reaction: The cell reaction for the galvanic

cell is:

2H+

(aq)+ 2e−→H2(g)E◦= 0 V

Ag+

(aq)+ e−→Ag(s)E◦= 0.80 V

Overall reaction:

2Ag+

(aq)+ 2H2(g)→H+

(aq)+ 2Ag(s)

Step 2: Write the Nernst equation: The Nernst equation is given by:

E=E◦−0.0592

nlog Q

K

where: - Eis the cell potential under non-standard conditions, - E◦is the

standard cell potential, - 0.0592 is the constant at 25◦C, - nis the number of

moles of electrons transferred in the cell reaction, - Qis the reaction quotient,

and - Kis the equilibrium constant.

Step 3: Calculate n, the number of moles of electrons transferred: From the

balanced overall reaction, we see that n= 2.

Step 4: Calculate Q, the reaction quotient:

Q=[H+]2

[Ag+]2

Given that [Ag+] = 1.0×10−3M and the concentration of H+ions is 1 M, we

have:

Q=(1.0)2

(1.0×10−3)2= 1.0×106

Step 5: Substitute values into the Nernst equation:

E= 0.80 V −0.0592

2log1.0×106

E= 0.80 V −0.0296 ×6

E= 0.80 V −0.1776 V

E= 0.6224 V

Therefore, the cell potential at 25◦C using the Nernst equation is 0.6224 V.

Question 18

Question

A concentration cell is set up using two half-cells, each containing a copper

electrode. One half-cell has a Cu2+ concentration of 0.10 M and the other half-

cell has a Cu2+ concentration of 0.0010 M. Calculate the cell potential at 25

using the Nernst equation.

Solution

Step 1: Write the half-reactions for the cell.

Anode: Cu(s)→Cu2+(aq)+2e−

Cathode: Cu2+(aq)+2e−→Cu(s)

Step 2: Write the cell reaction.

Cu(s)→Cu2+(0.10M, aq)|| Cu(s)←Cu2+(0.0010M, aq)

Step 3: Calculate the cell potential without considering concentrations. The

cell potential E◦

cell can be calculated using the standard reduction potentials for

the half-reactions. The standard reduction potential for the Cu reduction half-

reaction is +0.34 V. Since the two half-reactions are the same, the standard cell

potential is simply the difference between the two standard reduction potentials:

E◦

cell =E◦

cathode −E◦

anode = 0.34V−0.34V= 0V

Step 4: Calculate the cell potential with concentrations using the Nernst

equation. The Nernst equation is given by:

Ecell =E◦

cell −0.0592

nlog [Cu2+]cathode

[Cu2+]anode 

where - Ecell is the cell potential, - nis the number of electrons transferred in

the balanced cell reaction, - [Cu2+]cathode is the concentration of Cu2+ at the

cathode, and - [Cu2+]anode is the concentration of Cu2+ at the anode.

Since 2 moles of electrons are transferred in the reaction, n= 2.

Step 5: Substitute the values into the Nernst equation and calculate the cell

potential.

Ecell = 0 V −0.0592

2log 0.10

0.0010

Ecell = 0 V −0.0296 log(100)

Ecell = 0 V −0.0296 ×2

Ecell = 0 V −0.0592 V

Ecell =−0.0592 V

Therefore, the cell potential at 25

C is -0.0592 V.

Question 19

Question

Calculate the cell potential for a galvanic cell where the reaction is 2Cl−(aq)→

Cl2(g)+2e−and the following concentrations are present: [Cl−]=0.10 M,

[Cl2]=0.20 M. (Given: E◦

cell = 1.36 V, R= 8.314 J/(mol K), T= 298 K,

F= 96485 C/mol)

Solution

Step 1: Write the Nernst equation:

Ecell =E◦

cell −RT

nF ln(Q)

where Ecell is the cell potential, E◦

cell is the standard cell potential, Ris the gas

constant, Tis the temperature in Kelvin, nis the number of moles of electrons

transferred in the balanced redox reaction, Fis Faraday’s constant, and Qis

the reaction quotient.

Step 2: Determine the number of moles of electrons transferred (n) from the

balanced redox reaction. The balanced redox reaction is 2Cl−(aq)→Cl2(g) +

2e−, where 2 moles of electrons are transferred.

Step 3: Calculate the reaction quotient (Q) using the concentrations pro-

vided.

Q=[Cl2]2

[Cl−]2

Q=(0.20)2

(0.10)2= 4

Step 4: Substitute the given values into the Nernst equation.

Ecell = 1.36 −(8.314)(298)

(2)(96485) ln(4)

Step 5: Calculate the cell potential (Ecell).

Ecell = 1.36 −2470.312

192970 ln(4)

Ecell = 1.36 −0.0128

2.8826 ln(4)

Ecell = 1.36 −0.00004451 ln(4)

Ecell1.3599 V

Therefore, the cell potential for the given galvanic cell is approximately

1.3599 V.

Question 20

Question

A galvanic cell is constructed with a copper electrode in a 1.0 M CuSO4solution

and a silver electrode in a 1.0 M AgNO3solution. The standard reduction

potentials are E◦

Cu2+/Cu = 0.34 V and E◦

Ag+/Ag = 0.80 V, respectively. Calculate

the cell potential at 25

C for this cell.

Solution

Step 1: Write the half-reactions for the cell. The overall cell reaction can be

broken down into two half-reactions:

Cu2+ + 2e−→Cu (anode)

Ag++e−→Ag (cathode)

Step 2: Calculate the cell potential at standard conditions. Using the stan-

dard reduction potentials provided, we can calculate the standard cell potential,

E◦

cell, using the formula:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.80 V −0.34 V = 0.46 V

Step 3: Calculate the reaction quotient (Q). The reaction quotient, Q, can

be calculated using the concentrations of products and reactants:

Q=[Ag+]

[Cu2+]

Given that the solutions are 1.0 M, Q=1.0

1.0= 1.0

Step 4: Calculate the cell potential using the Nernst equation. The Nernst

equation relates the cell potential under non-standard conditions to the standard

cell potential:

Ecell =E◦

cell −0.0592

nlog(Q)

where nis the number of electrons transferred in the balanced equation. In this

case, n= 1. Plugging in the values, we get:

Ecell = 0.46 V −0.0592

1log(1.0)

Ecell = 0.46 V

Therefore, the cell potential at 25

C for this cell is 0.46 V.

Question 21

Question

A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper

electrode. The standard reduction potential of the copper electrode is E◦= 0.34

V. The concentration of copper ions in the cell is [Cu2+]=0.10 M. Calculate

the cell potential at 25

C using the Nernst equation.

Solution

Step 1: Write the half-reaction for the copper electrode. The half-reaction for

the copper electrode is:

Cu2+(aq)+2e−→Cu(s)

Step 2: Write the cell reaction and the cell potential. The cell reaction is:

2H++Cu2+ →H2+Cu

The standard cell potential (E◦

cell) can be calculated using the standard reduc-

tion potentials as follows: E◦

cell =E◦

cathode −E◦

anode =E◦

Cu −E◦

H+/H2= 0.34 V

−0 V = 0.34 V

Step 3: Determine the number of electrons transferred in the reaction. From

the cell reaction, we can see that 2 electrons are transferred.

Step 4: Write the Nernst equation. The Nernst equation is given by:

Ecell =E◦

cell −0.0592V

nlog [Cathode]

[Anode] 

where Ecell is the cell potential, E◦

cell is the standard cell potential, nis the num-

ber of electrons transferred, and [Cathode] and [Anode] are the concentrations

of the cathode and anode species, respectively.

Step 5: Calculate the cell potential. Substitute the given values into the

Nernst equation:

Ecell = 0.34 V −0.0592 V

2log 0.10

1

Ecell = 0.34 V −0.0296 log(0.10)

Ecell = 0.34 V −0.0296 ×(−1)

Ecell = 0.368 V

Therefore, the cell potential at 25

C using the Nernst equation is 0.368 V.

Question 22

Question

A voltaic cell consists of a silver electrode in a 1.0 M AgNO3solution and a

copper electrode in a 1.0 M CuSO4solution. Given that the standard reduction

potential for the Ag+(aq)/Ag(s) half-reaction is E◦= 0.80 V and the stan-

dard reduction potential for the Cu2+(aq)/Cu(s) half-reaction is E◦= 0.34 V,

calculate the cell potential at 25◦C.

Solution

Step 1: Write the two half-reactions for the voltaic cell:

Ag+(aq) + e−→Ag(s) E◦= 0.80 V

Cu2+(aq) + 2e−→Cu(s) E◦= 0.34 V

Step 2: Calculate the cell potential, Ecell, using the Nernst equation:

Ecell =E◦−0.0592

nlog [Ag+]

[Cu2+]2

Step 3: Determine the number of electrons transferred in the balanced re-

action. In this case, 2 moles of electrons are transferred in the reduction of 1

mole of Cu2+ ions.

Step 4: Calculate the cell potential at 25◦C:

Ecell = (0.80 V) −0.0592

2log 1.0 M

(1.0 M)2

= 0.80 V −0.0296 log(1.0)

= 0.80 V −0.0296(0)

= 0.80 V

Therefore, the cell potential at 25◦C is 0.80 V.

Question 23

Question

Calculate the cell potential for the following reaction at 25

2 AgCl(s)→2 Ag(s) + Cl2(g)

Given:

E◦

cell = 0.16 V

n= 2

Q= 1.0×10−5

Solution

Step 1: Write the Nernst equation

Ecell =E◦

cell −0.0592

nlog Q

Step 2: Substitute the given values into the Nernst equation

Ecell = 0.16 V −0.0592

2log1.0×10−5

Step 3: Calculate the cell potential

Ecell = 0.16 V −0.0296 log1.0×10−5

Ecell = 0.16 V −0.0296 ×(−5)

Ecell = 0.16 V + 0.148

Ecell = 0.308 V

Therefore, the cell potential for the given reaction at 25

C is 0.308 V.

Question 24

Question

A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper

electrode. The half-reaction occurring at the copper electrode is:

Cu2+ + 2e−→Cu(s)

The initial concentration of Cu2+ is 0.0100 M. At 25

C, the cell potential

is measured to be 0.66 V. Determine the concentration of Cu2+ when the cell

potential becomes 0.62 V.

Given: E◦

cell = 0.34 V R= 8.314 J/(mol·K) F= 96,485 C/mol

Solution

Step 1: Begin by writing the Nernst equation, which relates the cell potential

to the concentrations of species involved in the redox reaction:

Ecell =E◦

cell −RT

nF ln [Cu2+]

[Cu] 

Step 2: Substitute the known values into the equation and solve for [Cu2+]:

0.62V= 0.34V−(8.314 J/(mol ·K)) ·(298K)

2·(96485 C/mol) ln x

0.0100

Step 3: Simplify the equation:

0.62V= 0.34V−2473 J/mol

1932 C/mol ln x

0.0100

Step 4: Rearrange the equation and solve for x:

ln x

0.0100=(0.34 −0.62) V ·1932 C/mol

−2473 J/mol

Step 5: Calculate the value of x:

x= 0.0100 ×e





(0.34 −0.62) V ·1932 C/mol

−2473 J/mol





Step 6: Determine the concentration of Cu2+:

x= 0.0100 ×e





(0.34 −0.62) V ·1932 C/mol

−2473 J/mol



≈0.0036 M

Therefore, the concentration of Cu2+ when the cell potential becomes 0.62

V is approximately 0.0036 M.

Question 25

Question

A concentration cell is set up with two half-cells, each containing a Mn3+/Mn2+

half-reaction. One half-cell has a Mn3+ concentration of 0.10 M and the other

half-cell has a Mn3+ concentration of 0.0010 M. Calculate the cell potential

at 25

C for this concentration cell. The standard reduction potential for the

Mn3+/Mn2+ half-reaction is +1.57 V.

Solution

Step 1: Write the balanced half-reaction for the Mn3+/Mn2+ system. The

balanced half-reaction is:

Mn3+ + 1e−→Mn2+

Step 2: Calculate the cell potential using the Nernst equation: The Nernst

equation is given by:

E=E◦−RT

nF ln(Q)

Where: - E= cell potential - E◦= standard cell potential - R= gas constant

(8.314 J/(mol

K)) - T= temperature in Kelvin (25

C = 298 K) - n= number

of moles of electrons transferred in the balanced equation (1 in this case) - F=

Faraday’s constant (96485 C/mol) - Q= reaction quotient

Step 3: Calculate the reaction quotient Qfor the cell.

Q=[Mn2+]high

[Mn2+]low

Q=0.10

0.0010 = 100

Step 4: Substitute the given values into the Nernst equation and solve for

E= 1.57 V −(8.314 J/(mol

K))(298 K)

(1)(96485 C/mol) ln(100)

E= 1.57 V −2476.572

96485 ln(100) V

E≈1.27 V

Therefore, the cell potential at 25

C for this concentration cell is approxi-

mately 1.27 V.

Question 26

Question

Calculate the standard cell potential for the following reaction:

2 Fe3+(aq) + 2 I−(aq)→2 Fe2+(aq)+I2(s)

Given: E◦

Fe3+/Fe2+ = 0.77 V, [Fe3+]=0.010 M, [I−]=0.20 M and the standard

reduction potential for I2/2 I−is 0.54 V.

Solution

Step 1: Write the half-reactions and determine the cell potential.

Cathode: 2 Fe3+(aq) + 2 e−→2 Fe2+(aq)E◦

cathode = 0.77 V

Anode: 2 I−(aq)→I2(s) + 2 e−E◦

anode =−0.54 V

The cell potential is given by E◦

cell =E◦

cathode −E◦

anode. Substitute the given

values:

E◦

cell = 0.77 V −(−0.54 V) = 1.31 V

Step 2: Calculate the reaction quotient (Q) for the cell reaction. The reaction

quotient is given by:

Q=[Fe2+]2[I2]

[Fe3+]2[I−]2

Substitute the given concentrations:

Q=(0.010 M)2(1)

(0.20 M)2= 0.0025

Step 3: Use the Nernst equation to find the cell potential at non-standard

conditions. The Nernst equation is given by:

E=E◦−0.0592

nlog Q

where nis the number of moles of electrons transferred in the balanced redox

reaction. For the given cell reaction, n= 2. Substitute the values into the

Nernst equation:

E= 1.31 V −0.0592

2log 0.0025

E= 1.31 V −0.0298 log 0.0025

E≈1.31 V + 0.0872

E≈1.40 V

Therefore, the standard cell potential for the given reaction is 1.31 V and

the cell potential under the specified non-standard conditions is approximately

1.40 V.

Question 27

Question

A concentration cell is set up with two half-cells: one containing a zinc electrode

in a 0.10 M Zn(NO3)2solution and the other containing a zinc electrode in a

0.0010 M Zn(NO3)2solution. Calculate the cell potential for this concentration

cell at 25

C. The standard reduction potential for the Zn2+/Zn half-cell is -0.76

Solution

Step 1: Write the half-reactions for the oxidation and reduction half-cells. The

overall reaction for the cell involves the oxidation of Zn in the 0.10 M solution

and the reduction of Zn2+ ions in the 0.0010 M solution: Oxidation: Zn(s) →

Zn2+(aq) + 2e−Reduction: Zn2+(aq) + 2e−→Zn(s)

Step 2: Calculate the standard cell potential using the Nernst equation: E◦

= E◦

red - E◦

ox. E◦= 0.00 V - (-0.76 V) = 0.76 V

Step 3: Calculate the cell potential using the Nernst equation: E = E◦-

0.0592

nlog [Zn2+]low

[Zn2+]high . Here, n = 2 (from the balanced half-reactions), [Zn2+]low

= 0.0010 M, and [Zn2+]high = 0.10 M.

Step 4: Calculate the cell potential. E = 0.76 V - 0.0592

2log 0.0010

0.10 E =

0.76 V - 0.0296 log(0.01) E = 0.76 V - 0.0296 ×(-2) E = 0.76 V + 0.0592 E =

0.82 V

Therefore, the cell potential for this concentration cell at 25

C is 0.82 V.

Question 28

Question

A concentration cell is set up using two silver-silver chloride electrodes, one with

0.10 M AgCl

and the other with

1.0×10−3M AgCl

. If the standard reduction potential for the silver-silver chloride electrode is

+0.222 V

25◦C

, calculate the cell potential at

25◦C

Solution

Step 1: Write the overall cell reaction for the concentration cell. Label the

anode and cathode. The overall cell reaction for the concentration cell is:

AgCl(s) →Ag+(aq) + Cl−(aq)

In the cell with

0.10 M AgCl

acting as the anode and

1.0×10−3M AgCl

acting as the cathode, the anode is the electrode with higher concentration and

the cathode is the electrode with lower concentration.

Step 2: Write half-reactions for the anode and cathode, including the stan-

dard reduction potential. The half-reaction for the anode is:

AgCl(s) + e−→Ag+(aq) + Cl−(aq)

The half-reaction for the cathode is:

Ag+(aq)+e−→Ag(s)

Step 3: Calculate the cell potential at

25◦C

using the Nernst equation: The Nernst equation is given by:

E=E◦−0.0592

nlog [oxidized form]

[reduced form] 

For the anode half-reaction:

E◦

anode = +0.222 V

For the cathode half-reaction:

E◦

cathode = +0.222 V

The number of electrons transferred in each half-reaction (n) is 1.

Plugging the values into the Nernst equation for Ecell:

Ecell =Ecathode −Eanode =E◦

cathode −E◦

anode −0.0592

1log 1.0×10−3

0.10 

Ecell = 0.222 V −0.222 V −0.0592 log 1.0×10−3

0.10 

Ecell =−0.0592 log(0.01)

Ecell =−0.0592 × −2

Ecell = 0.1184 V

Therefore, the cell potential at

25◦C

for the concentration cell is

0.1184 V

Question 29

Question

A voltaic cell consists of a standard hydrogen electrode (SHE) in one half-cell

and a copper metal electrode in the other half-cell. The initial concentrations

are [Cu2+]=0.20 M and [H+]=1.0×10−3M. If the cell potential is measured

to be 0.45 V at 25◦C, calculate the standard reduction potential for the copper

half-reaction (Cu2+ + 2e−→Cu).

Solution

Step 1: Write the balanced equation for the overall reaction of the cell and the

half-reactions for the copper electrode: The overall cell reaction is:

2H++ Cu2+ →H2+ Cu

The half-reactions are: Cathode: Cu2+ + 2e−→Cu

Anode: 2H++ 2e−→H2

Step 2: Calculate the standard cell potential (E◦

cell) using the given reduction

potentials for the standard hydrogen electrode (E◦

SHE = 0 V) and the half-

reaction for the copper electrode. The standard cell potential is given by the

Nernst equation:

E◦

cell =E◦

cathode −E◦

anode

where E◦

cathode and E◦

anode are the standard reduction potentials for the cathode

and anode, respectively.

Given that E◦

cell = 0.45 V and E◦

SHE = 0 V, we can rewrite the Nernst

equation as:

0.45 V = E◦

Cu −0

E◦

Cu = 0.45 V

Therefore, the standard reduction potential for the copper half-reaction is

0.45 V.

Question 30

Question

A galvanic cell is constructed with silver metal and a silver ion solution, as well

as copper metal and a copper ion solution. The standard reduction potential

for the silver ion solution is E◦= 0.80 V and for the copper ion solution is

E◦= 0.34 V. The initial concentrations are [Ag+] = 1.0 M, [Cu2+] = 0.10 M.

Calculate the cell potential at 25

C after the cell has run for 15 minutes if

both solutions are at 25

C. Assume the exchange of ions is not limited by any

resistance.

Solution

Step 1: Write the overall cell reaction for the galvanic cell. The overall cell

reaction is the reduction of the silver ion solution and the oxidation of the

copper metal:

Ag+(aq)+e−−→ Ag(s)

Cu(s)−→ Cu2+(aq) + 2e−

Step 2: Calculate the cell potential at standard conditions. The standard

cell potential is given by the equation:

E◦

cell =E◦

cathode −E◦

anode

Substitute the given values:

E◦

cell = 0.80 V −0.34 V = 0.46 V

Step 3: Calculate the reaction quotient, Q. The reaction quotient Q at any

time t is given by:

Q=[Ag](t)

[Cu2+](t)2

Given that [Ag](t)=0.90 M and [Cu2+](t)=0.07 M, we can find Q.

Step 4: Calculate the cell potential at non-standard conditions using the

Nernst equation. The Nernst equation relates the standard cell potential to the

cell potential under non-standard conditions:

Ecell =E◦

cell −0.0592

nlog Q

K

Substitute the values of E◦

cell, Q, n, and K, and calculate Ecell.

Step 5: Calculate the cell potential after running for 15 minutes. Given

that the cell has run for 15 minutes, we can assume the system has reached

equilibrium and calculate the cell potential using Ecell obtained in the previous

step.

Question 31

Question

Calculate the cell potential (Ecell) for the following reaction at 25

Cu2+(0.010M) + Fe(s)→Cu(s) + Fe2+(1.0M)

Given the standard reduction potentials: E◦

Cu2+/Cu(s)= 0.34 V and E◦

Fe2+/Fe(s)=

−0.44 V.

Solution

Step 1: Write the half-reactions and their standard reduction potentials. The

half-reactions for the given reaction are:

Cu2+ + 2e−→Cu E◦

Cu2+/Cu(s)= 0.34 V

Fe2+ + 2e−→Fe E◦

Fe2+/Fe(s)=−0.44 V

Step 2: Calculate the cell potential using the Nernst equation:

The Nernst equation is given by:

Ecell =E◦

cell −0.0592

nlog Q

where Ecell is the cell potential, E◦

cell is the standard cell potential, nis the

number of electrons transferred in the balanced equation, and Qis the reaction

quotient.

Step 3: Calculate the standard cell potential (E◦

cell).

The standard cell potential is the difference between the standard reduction

potentials of the two half-reactions:

E◦

cell =E◦

cathode −E◦

anode

Plugging in the values:

E◦

cell = 0.34 V −(−0.44 V) = 0.78 V

Step 4: Calculate the reaction quotient (Q).

The reaction quotient is given by the concentration of products raised to

their stoichiometric coefficients divided by the concentration of reactants raised

to their stoichiometric coefficients:

Q=[Cu]α[Fe2+]β

[Cu2+]γ

In this case, α= 1, β= 1, and γ= 1.

Plugging in the concentrations:

Q=(1.0)(1.0)

(0.010) = 100

Step 5: Calculate the cell potential (Ecell).

Ecell = 0.78 V −0.0592

2log 100

Ecell = 0.78 V −0.0296 log 100

Ecell = 0.78 V −0.0296(2)

Ecell = 0.78 V −0.0592

Ecell = 0.7208 V

Therefore, the cell potential Ecell for the given reaction at 25

C is 0.7208 V.

Question 32

Question

A half-cell is set up with a silver electrode and a Ag+ion in a 0.50 M solution.

The half-cell is connected to a standard hydrogen electrode (Pt|H2(g, 1 atm)|H+(1 M))

through a salt bridge. Given that E0

Ag+/Ag = 0.80 V and the temperature is

25◦C, calculate the cell potential, Ecell, under standard conditions.

Solution

Step 1: Write the half-reactions for each electrode.

The half-reaction for the silver electrode is:

Ag++ e−→Ag

The half-reaction for the standard hydrogen electrode is:

H++ e−→1

2H2

Step 2: Write the cell reaction.

The cell reaction is the sum of the two half-reactions:

Ag++ H+→Ag + 1

2H2

Step 3: Calculate the cell potential under standard conditions using the

Nernst equation:

Ecell =E0

cell −0.0592

nlog [Ag][H+]

[Ag+][H2]1/2

Since the cell is set up with Ag and Ag+, the concentrations of Ag and Ag+

are equal in the equation. Therefore, the Nernst term simplifies to:

[H+]

[H2]1/2

Given that [H+] = 1 M for the standard hydrogen electrode, the Nernst term

becomes: 1

(1)1/2= 1

Thus, the cell potential under standard conditions is simply the standard

cell potential:

Ecell =E0

cell =E0

Ag+/Ag = 0.80 V

Therefore, the cell potential under standard conditions is 0.80 V.

Question 33

Question

A concentration cell is set up with two half-cells. In one half-cell, the concentra-

tion of Mn2+ is 0.10 M, and in the other half-cell, the concentration of Mn2+ is

1.0×10−3M. If E

for the reaction is E=−1.18 V, calculate the cell potential

at 298 K.

Solution

Step 1: Given the concentrations of Mn2+ ions and E

for the reaction, we can

use the Nernst equation to calculate the cell potential at 298 K. The Nernst

equation is given by:

E=E−0.0592

nlog [Mn2+]right

[Mn2+]left 

Where: - Eis the cell potential. - Eis the standard cell potential. - nis the

number of moles of electrons transferred in the balanced chemical equation. -

[Mn2+]right is the concentration of Mn2+ on the right half-cell. - [Mn2+]left is

the concentration of Mn2+ on the left half-cell.

Step 2: First, let’s calculate the number of moles of electrons transferred in

the balanced chemical equation. Since we don’t have the balanced equation, it

is generally equal to the absolute value of the stoichiometric coefficient of e−.

So n= 1 in this case.

Step 3: Plug in the values into the Nernst equation and solve for the cell

potential E:

E=−1.18 −0.0592

1log 1.0×10−3

0.10 

Step 4: Calculate the logarithmic term:

log 1.0×10−3

0.10 = log1.0×10−2=−2

Step 5: Substitute the calculated logarithmic term back into the Nernst

equation:

E=−1.18 −0.0592(−2)

Step 6: Calculate the cell potential E:

E=−1.18 + 0.1184 = −1.0616 V

Therefore, the cell potential at 298 K is −1.0616 V.

Question 34

Question

Calculate the cell potential at 25

C for a galvanic cell where the following half-

reactions are involved:

Zn2+ + 2e−→Zn(s) E◦=−0.76 V

Fe3+ +e−→Fe2+ E◦= 0.77 V

Given that the concentration of Zn2+ is 2.0×10−2M and the concentration of

Fe3+ is 1.0×10−4M.

Solution

Step 1: Write the overall cell reaction by adding the two half-reactions. The Fe

half-reaction needs to be multiplied by 2 to balance the electrons:

Zn2+ + Fe3+ →Zn(s) + Fe2+

Step 2: Determine the standard cell potential E◦

cell using the given standard

reduction potentials: E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.77 −(−0.76)

E◦

cell = 1.53 V

Step 3: Calculate the reaction quotient Qusing the concentrations provided:

Q=[Fe2+]

[Zn2+][Fe3+]=1.0×10−4

(2.0×10−2)(1.0×10−4)= 0.5

Step 4: Use the Nernst equation to calculate the cell potential Ecell at 25

Ecell =E◦

cell −0.0592

nlog Q

Where nis the number of electrons transferred in the balanced cell reaction.

Step 5: Since 2 moles of electrons are transferred in this reaction, substitute

n= 2 and Q= 0.5 into the Nernst equation:

Ecell = 1.53 −0.0592

2log(0.5)

Ecell = 1.53 −0.0296 ×(−0.30)

Ecell = 1.53 + 0.0089

Ecell = 1.5389 V

Therefore, the cell potential at 25

C for this galvanic cell is 1.5389 V.

Question 35

Question

A concentration cell is set up with two half-cells, one containing a 0.10 M solu-

tion of Fe2+ and the other containing a 0.0010 M solution of Fe2+. Calculate

the cell potential at 25

C for this concentration cell. Given that the stan-

dard reduction potential for Fe2+/Fe is +0.77 V and that the gas constant

R= 8.314 J/mol

Solution

Step 1: Write the half-reactions for the cell and find the overall cell reaction.

The half-reactions for the cell are:

Fe2+(0.10 M) + 2e−→Fe(s)E◦

cathode = +0.77 V

Fe2+(0.0010 M) + 2e−→Fe(s)E◦

anode = +0.77 V

The overall cell reaction is the combination of these two half-reactions:

Fe2+(0.10 M) →Fe2+(0.0010 M)

Step 2: Calculate the cell potential using the Nernst equation. The Nernst

equation relates the cell potential to the standard cell potential and the concen-

trations of the reactants in a cell:

E=E◦−0.0592

nlog [Fe2+]anode

[Fe2+]cathode 

Where: - Eis the cell potential, - E◦is the standard cell potential, - nis

the number of moles of electrons transferred in the balanced equation (in this

case, n= 2), - [Fe2+]anode is the concentration of Fe2+ at the anode, and -

[Fe2+]cathode is the concentration of Fe2+ at the cathode.

Plugging in the values:

E= +0.77 V −0.0592

2log 0.0010

0.10 

E= +0.77 V + 0.0296 log(0.01)

E= +0.77 V + 0.0296(−2)

E= +0.77 V −0.0592

E= +0.7108 V

Therefore, the cell potential at 25

C for this concentration cell is +0.7108 V .

Substitute the given values into the equation:

E= 1.51 V −0.0592

5log 0.10

(1.0×10−5)5(1)8

E= 1.51 V −0.0592

5log 0.10

1.0×10−25 

E= 1.51 V −0.0592

5log1025

E= 1.51 V −0.0592

5×25

E= 1.51 V −0.1184

E= 1.3916 V

Therefore, the cell potential of the galvanic cell at 25

C is 1.3916 V.

Question 2

Question

Calculate the cell potential for the following redox reaction at 25

Zn(s) + F e2+(aq)→Zn2+(aq) + F e(s)

given that [F e2+] = 0.10 M, [Zn2+]=1.0 M, and the standard reduction poten-

tials are E0

Fe2+/Fe =−0.44 V and E0

Zn2+/Zn =−0.76 V.

Solution

Step 1: Write the half-reactions and the overall cell reaction. The half-reactions

are:

Cathode (Reduction): F e2+(aq)+2e−→F e(s)E0

cathode =−0.44 V

Anode (Oxidation): Zn(s)→Zn2+(aq)+2e−E0

anode =−0.76 V

The overall reaction is the sum of the two half-reactions:

Zn(s) + F e2+(aq)→Zn2+(aq) + F e(s)

Step 2: Calculate the cell potential using the Nernst equation: The Nernst

equation is given by:

Ecell =E0

cell −0.0592

nlog Q

where: - Ecell is the cell potential, - E0

cell is the standard cell potential, - nis the

number of electrons transferred in the balanced cell reaction, - Qis the reaction

quotient, and - Kis the equilibrium constant.

For this reaction, n= 2 as 2 electrons are transferred. The reaction quotient

Qfor this reaction is:

Q=[Zn2+]

[F e2+]=1.0

0.10 = 10

Substitute the values into the Nernst equation:

Ecell = (−0.76 V) −0.0592

2log(10)

Ecell =−0.76 V −0.0296 log(10)

Ecell =−0.76 V −0.0296 ×1

Ecell =−0.76 V −0.0296

Ecell =−0.7896 V

Therefore, the cell potential for the given redox reaction at 25

C is −0.7896 V.

Question 3

Question

A galvanic cell contains a copper electrode in a 1.0 M solution of Cu2+ ions and

a silver electrode in a 0.10 M solution of Ag+ions. Calculate the cell potential

at 25

C given that E◦

cell = 0.46 V.

Solution

Step 1: Write the half-reactions for the cathode and anode. The reduction half-

reactions for the copper electrode (Cu2+ gaining electrons) and the silver elec-

trode (Ag+gaining electrons) are: Cathode: Cu2+(aq)+2e−→Cu(s)E◦

cathode =

0.34 V Anode: Ag+(aq) + e−→Ag(s)E◦

anode = 0.80 V

Step 2: Calculate the cell potential using the Nernst equation. The Nernst

equation is given by:

Ecell =E◦

cell −0.0592

nlog [reduced form (cathode)]m

[oxidized form (anode)]n

where: n= number of electrons transferred (n= 1 in this case) m= stoichio-

metric coefficient of the reduced form (cathode)

Ecell =E◦

cell −0.0592

1log [Cu(s)]

[Ag(s)] 

Step 3: Calculate the concentration of Cu2+ and Ag+. Given: [Cu2+] =

1.0 M and [Ag+] = 0.10 M

Step 4: Substitute the concentrations into the Nernst equation and solve for

Ecell.

Ecell = 0.46 V −0.0592

1log 1

0.10

Ecell = 0.46 V −0.0592 ×log(10)

Ecell = 0.46 V −0.0592 ×1

Ecell = 0.46 V −0.0592 V = 0.40 V

Therefore, the cell potential at 25

C is 0.40 V.

Question 4

Question

The concentration of Br−ions in a solution is 1.5×10−2M. The standard

reduction potential of the Br2/Br−half-reaction is +1.07 V. Calculate the cell

potential when the concentration of Br−ions is diluted to 5.0×10−3M at 25

(Assume the temperature coefficient αis 0.0 V/

C.)

Solution

Step 1: Write the Nernst equation. The Nernst equation relates the cell potential

to the concentrations of the reactants and products in an electrochemical cell:

E=E◦−0.0592

nlog [Reductant]

[Oxidant]

where: - Eis the cell potential, - E◦is the standard reduction potential, - nis

the number of electrons transferred in the half-reaction, and - [Reductant] and

[Oxidant] are the concentrations of the reductant and oxidant, respectively.

Step 2: Determine the number of electrons transferred, n. Since the half-

reaction involves the reaction Br2+2e−→2Br−, the number of electrons trans-

ferred is 2.

Step 3: Calculate the cell potential at the new concentration. First, calculate

the new cell potential using the Nernst equation:

E= 1.07 V −0.0592

2log 1.5×10−2

5.0×10−3

Step 4: Substitute the values and solve for E.

E= 1.07 V −0.0592

2log 1.5×10−2

5.0×10−3= 1.07 V −0.0592

2log 3

E= 1.07 V −0.0296 log 3

Step 5: Calculate the final cell potential.

E= 1.07 V −0.0296 ×0.4771 = 1.07 V −0.0141 = 1.0559 V

Therefore, the cell potential when the concentration of Br−ions is diluted

to 5.0×10−3M at 25

C is 1.0559 V.

Question 5

Question

A concentration cell is set up with two half-cells. One half-cell has a silver

electrode in a 0.10 M AgNO3solution, and the other half-cell has a silver

electrode in a 0.0010 M AgNO3solution. Calculate the cell potential for this

concentration cell at 298 K. (Given: E◦

cell = 0.80 V)

Solution

Step 1: Write the half-reactions involved in the concentration cell. The two half-

reactions involved are: Anode: Ag(s)→Ag+(aq) + e−(oxidation) Cathode:

Ag+(aq) + e−→Ag(s) (reduction)

Step 2: Write the Nernst equation for the cell potential (Ecell). The Nernst

equation is given by: Ecell =E◦

cell −0.0592

nlog [Ag+]anode

[Ag+]cathode 

Step 3: Calculate the concentration ratio of [Ag+]anode to [Ag+]cathode.

Given: [Ag+]anode = 0.10 M [Ag+]cathode = 0.0010 M

Concentration ratio: [Ag+]anode

[Ag+]cathode =0.10

0.0010 = 100

Step 4: Substitute the values into the Nernst equation and solve for Ecell.

Ecell = 0.80V−0.0592

1log(100) Ecell = 0.80V−0.0592×2Ecell = 0.80V−0.1184

Ecell = 0.6816 V

Therefore, the cell potential for this concentration cell at 298 K is 0.6816 V.

Question 6

Question

Given the following cell notation representing a galvanic cell:

Mn3+(aq)|Mn2+(aq)|| Cu2+(aq)|Cu(s)

where the reduction potentials are as follows: E0(Mn3+/Mn2+) = −1.18 V

and E0(Cu2+/Cu) = +0.34 V. Determine the cell potential at 25◦Cwhen

[Mn3+]=0.10 M, [Mn2+]=0.20 M, and [Cu2+] = 0.50 M.

Solution

Step 1: Write down the half-cell reactions for the given redox reactions. The

half-cell reactions are:

Cathode: Cu2+(aq)+2e−→Cu(s)

Anode: Mn3+(aq) + e−→Mn2+(aq)

Step 2: Calculate the standard cell potential, E0

cell, using the Nernst equa-

tion:

cell =E0

cathode −E0

anode

cell = (+0.34 V)−(−1.18 V) = 1.52 V

Step 3: Calculate the reaction quotient, Q, using the concentrations given:

Q=[Mn2+]

[Mn3+]×1

[Cu2+]=0.20

0.10 ×1

0.50 = 4.00

Step 4: Calculate the cell potential at 25◦Cusing the Nernst equation:

Ecell =E0

cell −0.0592 V

nlog(Q)

Since the reaction quotient, Q, is 4 and there are 1 electron transferred in this

reaction, we have:

Ecell = 1.52 V−0.0592 V

1log(4)

Ecell = 1.52 V−0.0592 V×0.6021

Ecell = 1.52 V−0.0358 V

Ecell = 1.4842 V

Therefore, the cell potential at 25◦Cis 1.4842 V.

Question 7

Question

Calculate the standard cell potential for the following reaction at 25

Zn(s) + 2Ag+(aq)→Zn2+(aq) + 2Ag(s)

Given:

E◦

Ag+/Ag = 0.80 V

E◦

Zn2+/Zn =−0.76 V

Solution

Step 1: Write the Nernst equation for the cell potential:

E=E◦−RT

nF ln(Q)

Step 2: Calculate the standard cell potential (E◦)

E◦=E◦

cathode −E◦

anode

E◦=E◦

Ag+/Ag −E◦

Zn2+/Zn

E◦= 0.80 V −(−0.76 V)

E◦= 1.56 V

Step 3: Calculate the reaction quotient (Q) for the given cell:

Q=[Zn2+]

[Ag+]2

Step 4: Substitute the values into the Nernst equation and calculate the cell

potential (E):

E= 1.56 V −(8.314J/Kmol)(298K)

2(96485C/mol)ln(Q)

E= 1.56 V −2482.452

1930.7ln(Q)

E= 1.56 V −1.2869 ln(Q)

Therefore, the cell potential for the given reaction at 25

C is given by E=

1.56 V −1.2869 ln(Q).

Question 8

Question

Calculate the cell potential for a galvanic cell in which [Cu2+] = 0.10 M, [Zn2+]

= 1.0 ×10−4M, [Cu] = 1.0 M, and [Zn] = 1.0 M, at 25

C. The standard

reduction potentials are E◦(Cu2+/Cu) = 0.34 V and E◦(Zn2+/Zn) = -0.76 V.

Solution

1. The cell potential at non-standard conditions is calculated using the Nernst

equation:

E = E◦−0.0592

nlog(Q)

where E is the cell potential, E◦is the standard cell potential, n is the

number of moles of electrons transferred, and Q is the reaction quotient.

2. The reaction taking place in the cell is: Zn(s) + Cu2+(aq) →Zn2+(aq) +

Cu(s).

3. The number of moles of electrons transferred in this reaction is 2, because

zinc loses 2 electrons and copper gains 2 electrons.

4. First, calculate the reaction quotient Q:

Q=[Zn2+][Cu]

[Zn][Cu2+]

Q=(1.0×10−4)(1.0)

(1.0)(0.10)

Q= 1.0×10−3

5. Now, substitute the given values into the Nernst equation:

E = 0.34 −0.0592

2log1.0×10−3

E = 0.34 −0.0296 ×3

E=0.34 −0.0888

E=0.2512 V

6. Therefore, the cell potential for the galvanic cell at these non-standard

conditions is 0.2512 V.

Question 9

Question

A concentration cell is constructed using two half-cells. One half-cell has a

silver wire in a silver ion solution with a concentration of 1.0×10−3M, and

the other half-cell has a silver wire in a silver ion solution with a concentration

of 1.0×10−2M. What is the cell potential at 25

C for this concentration cell?

Given that the standard reduction potential for the Ag+—Ag half-cell is 0.80

Solution

Step 1: Write the half-reactions for each half-cell. The two half-reactions in-

volved are:

Ag++e−→Ag (s)

Step 2: Write the Nernst equation for the cell potential. The Nernst equation

is given by:

Ecell =E◦

cell −0.0592

nlog [Ag+]cathode

[Ag+]anode 

where: Ecell = cell potential E◦

cell = standard cell potential n= number of moles

of electrons transferred (1 in this case) [Ag+]cathode = concentration of silver

ion in the cathode half-cell [Ag+]anode = concentration of silver ion in the anode

half-cell

Step 3: Calculate the cell potential. Substitute the given values into the

Nernst equation:

Ecell = 0.80 V −0.0592

1log 1.0×10−3M

1.0×10−2M

Ecell = 0.80 V −0.0592

1log(0.1)

Ecell = 0.80 V −0.0592 log(0.1)

Ecell = 0.80 V −0.0592(−1)

Ecell = 0.80 + 0.0592

Ecell = 0.8592 V

Therefore, the cell potential at 25

C for this concentration cell is 0.8592 V.

Question 10

Question

A cell consists of a zinc electrode in a 2.0 M Zn2+ solution and a copper electrode

in a 0.10 M Cu2+ solution. If the standard reduction potentials are E◦

Zn2+/Zn =

−0.76Vand E◦

Cu2+/Cu = +0.34V, calculate the cell potential at 25

C using the

Nernst equation.

Solution

Let’s use the Nernst equation to calculate the cell potential at 25

E=E◦−0.0592

nlog [Cathode]

[Anode] 

where: E= cell potential E◦= standard cell potential n= number of

electrons transferred [Cathode] = concentration of the cathode species [Anode]

= concentration of the anode species

First, calculate the number of electrons transferred (n) in the reaction. The

half-reactions are:

Zn2+ + 2e−→Zn E◦=−0.76

Cu2+ + 2e−→Cu E◦= +0.34

From the half-reactions, we can see that 2 electrons are transferred in each

reaction.

Step 1: Calculate the cell potential at 25

C for the given concentrations

using the Nernst equation:

E=E◦

cell −0.0592

2log [Cu2+]

[Zn2+]

Step 2: Substitute the given values into the equation:

E= 0.34V−0.0592

2log 0.10

2.0

E= 0.34V−0.0296 log(0.05)

Step 3: Calculate the natural logarithm:

E= 0.34V−0.0296 ×(−2.9957)

E= 0.34V+ 0.0887

Step 4: Calculate the cell potential:

E= 0.43V

Therefore, the cell potential at 25

C using the Nernst equation is 0.43 V.

Question 11

Question

A galvanic cell consists of a silver electrode in contact with a silver ion solution of

unknown concentration, and a zinc electrode in contact with a zinc ion solution

of concentration 0.10 M. The cell potential is measured to be +0.42 Vat 25◦C.

Determine the concentration of silver ions in the silver half-cell.

Solution

Step 1: Write the half-reactions for the cell. The cell consists of the following

two half-reactions: 1. Anode (Zinc electrode): Zn(s)→Zn2+(aq) + 2e−2.

Cathode (Silver electrode): Ag+(aq) + e−→Ag(s)

Step 2: Calculate cell potential using the Nernst equation. The Nernst

equation is given by:

E=E◦−RT

nF ln(Q)

Given that the cell potential Eis +0.42 V, the standard cell potential E◦is

+1.10 Vfor the reduction of Ag+ions to Ag(s). The temperature Tis 25◦C

which is equivalent to 298 K. The number of electrons transferred nin the

reaction is 1, the Faraday constant Fis 96485 C/mol, and the gas constant

Ris 8.314 J/(mol ·K). The reaction quotient Qcan be calculated using the

concentrations of Zn2+ and Ag+ions.

Step 3: Determine the concentration of Ag+ions in the silver half-cell.

To determine the concentration of Ag+ions, we first need to calculate the

value of the reaction quotient Q. Since the reaction is at equilibrium, Q=K

(equilibrium constant) for the cell reaction. The equilibrium constant expression

is given by:

K=[Ag(s)]

[Ag+(aq)]

Since the activity of a pure solid is considered to be 1, the equilibrium con-

stant can be simplified to 1

[Ag+(aq)] . Thus, Q=1

[Ag+(aq)] . Next, substitute the

values into the Nernst equation and solve for [Ag+(aq)]. This will give us the

concentration of silver ions in the silver half-cell.

Question 12

Question

A galvanic cell is constructed with a standard hydrogen electrode (SHE) as the

anode and a silver-silver chloride electrode as the cathode. The concentration

ceAg+ ions in the cathode compartment is 0.10 M and the partial pressure of

ceH2 gas in the anode compartment is 0.50 atm. Calculate the cell potential at

Given: Standard reduction potentials:

ceAg + +e− − > Ag(s)E0= 0.80V

ce2H+ +2e− − > H2(g)E0= 0.00V

Faraday’s constant, F= 96,485 C mol−1, Gas constant, R= 8.314 J K−1mol−1,

Temperature, T= 298 K

Solution

Step 1: Write the half-reactions for the anode and cathode: Anode (oxidation

at SHE):

ceH2(g)−>2H+ +2e−

Cathode (reduction at Ag/AgCl electrode):

ceAg + +e− − > Ag(s)

Step 2: Calculate the cell potential without considering non-standard con-

ditions:

cell =E0

cathode −E0

anode

cell = 0.80 V−0.00 V= 0.80 V

Step 3: Calculate the cell potential under non-standard conditions using the

Nernst equation:

Ecell =E0

cell −RT

nF ln Q

P

Where: Ecell = cell potential under non-standard conditions E0

cell = cell

potential under standard conditions R= gas constant T= temperature n=

number of moles of electrons transferred F= Faraday’s constant Q= reaction

quotient P= atmospheric pressure

Here, n= 2 (moles of electrons transferred)

Ecell = 0.80 V−(8.314 J K−1mol−1)(298 K)

(2)(96485 C mol−1)ln 0.10

0.50

Ecell = 0.80 V−(2476.52)

(2)(96485) ln(0.20) = 0.80 V−(0.0128) ln(0.20)

Ecell ≈0.80 V−0.0128(−0.693) = 0.80 V+ 0.0089 ≈0.81 V

Therefore, the cell potential at 25

C under these non-standard conditions is

approximately 0.81 V.

Question 13

Question

A voltaic cell is set up where the standard reduction potential of the cathode is

+0.42 V and that of the anode is -0.80 V. Calculate the cell potential when the

concentration of Mn2+ is 0.010 M at the anode and the concentration of Ag+

is 2.0 M at the cathode.

Solution

Step 1: Write the half-reactions and their standard reduction potentials. The

half-reactions involved are:

Anode: Mn2+(aq) →Mn(s) + 2e−E0=−0.80 V

Cathode: Ag+(aq) + e−→Ag(s) E0= +0.42 V

Step 2: Calculate the cell potential at standard conditions. The cell potential

at standard conditions is given by the formula:

cell =E0

cathode −E0

anode

cell = (+0.42 V) −(−0.80 V) = 1.22 V

Step 3: Determine the reaction quotient Q. The reaction quotient Q is given

by:

Q=[Mn]

[Ag+]=0.010

2.0= 0.0050

Step 4: Calculate the cell potential at non-standard conditions using the

Nernst equation. The Nernst equation is given by:

Ecell =E0

cell −0.0592

nlog(Q)

where n is the number of electrons transferred in the balanced redox reaction.

Step 5: Calculate the cell potential at non-standard conditions. Since 2

moles of electrons are transferred,

Ecell = 1.22 V −0.0592

2log(0.0050) = 1.22 V −0.0592 ×1×(−2.30) = 1.35 V

Therefore, the cell potential at non-standard conditions is 1.35 V.

Question 14

Question

At 25

C, a cell has a standard cell potential of 1.23 V and an [Cu2+] concentra-

tion of 0.10 M in the cathode compartment while the [Cu2+] concentration in the

anode compartment is 0.001 M. Calculate the cell potential at this nonstandard

condition.

Solution

The Nernst equation is given by:

E=E◦−0.0592

nlog Q

where: - Eis the cell potential at nonstandard conditions (in V) - E◦is the

standard cell potential (in V) - nis the number of moles of electrons transferred

in the balanced cell reaction - Qis the reaction quotient

Step 1: Write the balanced redox reaction for the cell. The cell reaction is:

Cu2+ + 2e−→Cu

Therefore, the number of moles of electrons transferred, n, is 2.

Step 2: Calculate the reaction quotient Q.

Q=[Cu]

[Cu2+]=0.10

0.001 = 100

Step 3: Substitute the known values into the Nernst equation.

E= 1.23 −0.0592

2log 100

Step 4: Simplify the equation.

E= 1.23 −0.0296 log 100 = 1.23 −0.0296 ×2=1.23 −0.0592 = 1.17 V

Therefore, the cell potential at this nonstandard condition is 1.17 V.

Question 15

Question

A concentration cell is set up using two half-cells, both containing standard

hydrogen electrodes. One half-cell has a hydrogen ion concentration of 0.1 M,

while the other half-cell has a hydrogen ion concentration of 0.01 M. Calculate

the cell potential at 25

C using the Nernst equation.

Solution

Step 1: Write the half-reactions and the cell reaction. The half-reactions for the

standard hydrogen electrode are:

Reduction: 2H++ 2e−→H2E◦

red = 0 V

Oxidation: H2→2H++ 2e−E◦

ox = 0 V

The overall cell reaction is the difference between the two half-reactions:

H2(0.1 M) →H2(0.01 M)

Step 2: Calculate the standard cell potential (E◦

cell). Since the standard

reduction potential of the standard hydrogen electrode is 0 V, the standard cell

potential is also 0 V.

Step 3: Calculate the reaction quotient (Q). The reaction quotient, Q, can

be calculated using the concentrations of the hydrogen ions in each half-cell:

Q=[H+

[H+

2]=0.1

0.01 = 10

Step 4: Calculate the cell potential at 25

C using the Nernst equation. The

Nernst equation relates the cell potential (Ecell) to the standard cell potential

(E◦

cell), the gas constant (R), the temperature (T), the number of electrons

transferred (n), and the Faraday constant (F):

Ecell =E◦

cell −RT

nF ln(Q)

Plugging in the values:

Ecell = 0 V −(8.314 J/mol ·K)(298 K)

2(96485 C/mol) ln(10)

Ecell = 0 V −8.314 J/mol ·K·298 K

2·96485 C/mol ln(10)

Ecell = 0 V −6221.252

192970 ln(10)

Ecell ≈ −0.019 V

Therefore, the cell potential at 25

C using the Nernst equation is approxi-

mately -0.019 V.

Question 16

Question

A voltaic cell consists of a standard hydrogen electrode (SHE) and a copper-

silver cell. The concentration of copper ions in the copper half-cell is 0.10 M,

while the concentration of silver ions in the silver half-cell is 1.0 M. Calculate

the cell potential at 25◦C when the copper electrode is the anode. Given:

E◦(Cu2+/Cu) = 0.34 V and E◦(Ag+/Ag) = 0.80 V.

Solution

Step 1: Write the cell reaction and determine the overall cell potential.

The cell reaction is:

Cu(s) + 2Ag+(aq)→Cu2+(aq) + 2Ag(s)

The overall cell potential (E◦

cell) can be calculated as:

E◦

cell =E◦

cathode −E◦

anode

Given E◦(Cu2+/Cu) = 0.34 V and E◦(Ag+/Ag) = 0.80 V, we have:

E◦

cell =E◦

Ag+/Ag −E◦

Cu2+/Cu

E◦

cell = 0.80 V −0.34 V

E◦

cell = 0.46 V

Therefore, the standard cell potential is 0.46 V.

Step 2: Apply the Nernst equation to calculate the cell potential at non-

standard conditions.

The Nernst equation is given by:

E=E◦−0.0592

nlog Q

Where: - Eis the cell potential under non-standard conditions. - E◦is the

standard cell potential. - nis the number of moles of electrons transferred in

the balanced redox reaction. - Qis the reaction quotient. - Kis the equilibrium

constant.

In this case, n= 2 (from the balanced reaction), K= 1.8×1010, and

Q= [Cu2+]/[Ag+]2.

Step 3: Calculate Q.

Q=0.10

(1.0)2= 0.10

Step 4: Substitute all known values into the Nernst equation and solve for

E= 0.46 V −0.0592

2log 0.10

1.02

E= 0.46 V −0.0296 log 0.10

E= 0.46 V −0.0296 ×(−1)

E= 0.46 V + 0.0296

E= 0.4896 V

Therefore, the cell potential at 25◦C when the copper electrode is the anode

is 0.4896 V.

Question 17

Question

A galvanic cell is set up with a standard hydrogen electrode (SHE) on one side

and a silver electrode on the other side. The standard reduction potential of the

silver electrode is E◦= 0.80 V. If the concentration of Ag+ions is 1.0×10−3M

in the solution, calculate the cell potential at 25◦C using the Nernst equation.

Solution

Step 1: Write down the standard cell reaction: The cell reaction for the galvanic

cell is:

2H+

(aq)+ 2e−→H2(g)E◦= 0 V

Ag+

(aq)+ e−→Ag(s)E◦= 0.80 V

Overall reaction:

2Ag+

(aq)+ 2H2(g)→H+

(aq)+ 2Ag(s)

Step 2: Write the Nernst equation: The Nernst equation is given by:

E=E◦−0.0592

nlog Q

K

where: - Eis the cell potential under non-standard conditions, - E◦is the

standard cell potential, - 0.0592 is the constant at 25◦C, - nis the number of

moles of electrons transferred in the cell reaction, - Qis the reaction quotient,

and - Kis the equilibrium constant.

Step 3: Calculate n, the number of moles of electrons transferred: From the

balanced overall reaction, we see that n= 2.

Step 4: Calculate Q, the reaction quotient:

Q=[H+]2

[Ag+]2

Given that [Ag+] = 1.0×10−3M and the concentration of H+ions is 1 M, we

have:

Q=(1.0)2

(1.0×10−3)2= 1.0×106

Step 5: Substitute values into the Nernst equation:

E= 0.80 V −0.0592

2log1.0×106

E= 0.80 V −0.0296 ×6

E= 0.80 V −0.1776 V

E= 0.6224 V

Therefore, the cell potential at 25◦C using the Nernst equation is 0.6224 V.

Question 18

Question

A concentration cell is set up using two half-cells, each containing a copper

electrode. One half-cell has a Cu2+ concentration of 0.10 M and the other half-

cell has a Cu2+ concentration of 0.0010 M. Calculate the cell potential at 25

using the Nernst equation.

Solution

Step 1: Write the half-reactions for the cell.

Anode: Cu(s)→Cu2+(aq)+2e−

Cathode: Cu2+(aq)+2e−→Cu(s)

Step 2: Write the cell reaction.

Cu(s)→Cu2+(0.10M, aq)|| Cu(s)←Cu2+(0.0010M, aq)

Step 3: Calculate the cell potential without considering concentrations. The

cell potential E◦

cell can be calculated using the standard reduction potentials for

the half-reactions. The standard reduction potential for the Cu reduction half-

reaction is +0.34 V. Since the two half-reactions are the same, the standard cell

potential is simply the difference between the two standard reduction potentials:

E◦

cell =E◦

cathode −E◦

anode = 0.34V−0.34V= 0V

Step 4: Calculate the cell potential with concentrations using the Nernst

equation. The Nernst equation is given by:

Ecell =E◦

cell −0.0592

nlog [Cu2+]cathode

[Cu2+]anode 

where - Ecell is the cell potential, - nis the number of electrons transferred in

the balanced cell reaction, - [Cu2+]cathode is the concentration of Cu2+ at the

cathode, and - [Cu2+]anode is the concentration of Cu2+ at the anode.

Since 2 moles of electrons are transferred in the reaction, n= 2.

Step 5: Substitute the values into the Nernst equation and calculate the cell

potential.

Ecell = 0 V −0.0592

2log 0.10

0.0010

Ecell = 0 V −0.0296 log(100)

Ecell = 0 V −0.0296 ×2

Ecell = 0 V −0.0592 V

Ecell =−0.0592 V

Therefore, the cell potential at 25

C is -0.0592 V.

Question 19

Question

Calculate the cell potential for a galvanic cell where the reaction is 2Cl−(aq)→

Cl2(g)+2e−and the following concentrations are present: [Cl−]=0.10 M,

[Cl2]=0.20 M. (Given: E◦

cell = 1.36 V, R= 8.314 J/(mol K), T= 298 K,

F= 96485 C/mol)

Solution

Step 1: Write the Nernst equation:

Ecell =E◦

cell −RT

nF ln(Q)

where Ecell is the cell potential, E◦

cell is the standard cell potential, Ris the gas

constant, Tis the temperature in Kelvin, nis the number of moles of electrons

transferred in the balanced redox reaction, Fis Faraday’s constant, and Qis

the reaction quotient.

Step 2: Determine the number of moles of electrons transferred (n) from the

balanced redox reaction. The balanced redox reaction is 2Cl−(aq)→Cl2(g) +

2e−, where 2 moles of electrons are transferred.

Step 3: Calculate the reaction quotient (Q) using the concentrations pro-

vided.

Q=[Cl2]2

[Cl−]2

Q=(0.20)2

(0.10)2= 4

Step 4: Substitute the given values into the Nernst equation.

Ecell = 1.36 −(8.314)(298)

(2)(96485) ln(4)

Step 5: Calculate the cell potential (Ecell).

Ecell = 1.36 −2470.312

192970 ln(4)

Ecell = 1.36 −0.0128

2.8826 ln(4)

Ecell = 1.36 −0.00004451 ln(4)

Ecell1.3599 V

Therefore, the cell potential for the given galvanic cell is approximately

1.3599 V.

Question 20

Question

A galvanic cell is constructed with a copper electrode in a 1.0 M CuSO4solution

and a silver electrode in a 1.0 M AgNO3solution. The standard reduction

potentials are E◦

Cu2+/Cu = 0.34 V and E◦

Ag+/Ag = 0.80 V, respectively. Calculate

the cell potential at 25

C for this cell.

Solution

Step 1: Write the half-reactions for the cell. The overall cell reaction can be

broken down into two half-reactions:

Cu2+ + 2e−→Cu (anode)

Ag++e−→Ag (cathode)

Step 2: Calculate the cell potential at standard conditions. Using the stan-

dard reduction potentials provided, we can calculate the standard cell potential,

E◦

cell, using the formula:

E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.80 V −0.34 V = 0.46 V

Step 3: Calculate the reaction quotient (Q). The reaction quotient, Q, can

be calculated using the concentrations of products and reactants:

Q=[Ag+]

[Cu2+]

Given that the solutions are 1.0 M, Q=1.0

1.0= 1.0

Step 4: Calculate the cell potential using the Nernst equation. The Nernst

equation relates the cell potential under non-standard conditions to the standard

cell potential:

Ecell =E◦

cell −0.0592

nlog(Q)

where nis the number of electrons transferred in the balanced equation. In this

case, n= 1. Plugging in the values, we get:

Ecell = 0.46 V −0.0592

1log(1.0)

Ecell = 0.46 V

Therefore, the cell potential at 25

C for this cell is 0.46 V.

Question 21

Question

A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper

electrode. The standard reduction potential of the copper electrode is E◦= 0.34

V. The concentration of copper ions in the cell is [Cu2+]=0.10 M. Calculate

the cell potential at 25

C using the Nernst equation.

Solution

Step 1: Write the half-reaction for the copper electrode. The half-reaction for

the copper electrode is:

Cu2+(aq)+2e−→Cu(s)

Step 2: Write the cell reaction and the cell potential. The cell reaction is:

2H++Cu2+ →H2+Cu

The standard cell potential (E◦

cell) can be calculated using the standard reduc-

tion potentials as follows: E◦

cell =E◦

cathode −E◦

anode =E◦

Cu −E◦

H+/H2= 0.34 V

−0 V = 0.34 V

Step 3: Determine the number of electrons transferred in the reaction. From

the cell reaction, we can see that 2 electrons are transferred.

Step 4: Write the Nernst equation. The Nernst equation is given by:

Ecell =E◦

cell −0.0592V

nlog [Cathode]

[Anode] 

where Ecell is the cell potential, E◦

cell is the standard cell potential, nis the num-

ber of electrons transferred, and [Cathode] and [Anode] are the concentrations

of the cathode and anode species, respectively.

Step 5: Calculate the cell potential. Substitute the given values into the

Nernst equation:

Ecell = 0.34 V −0.0592 V

2log 0.10

1

Ecell = 0.34 V −0.0296 log(0.10)

Ecell = 0.34 V −0.0296 ×(−1)

Ecell = 0.368 V

Therefore, the cell potential at 25

C using the Nernst equation is 0.368 V.

Question 22

Question

A voltaic cell consists of a silver electrode in a 1.0 M AgNO3solution and a

copper electrode in a 1.0 M CuSO4solution. Given that the standard reduction

potential for the Ag+(aq)/Ag(s) half-reaction is E◦= 0.80 V and the stan-

dard reduction potential for the Cu2+(aq)/Cu(s) half-reaction is E◦= 0.34 V,

calculate the cell potential at 25◦C.

Solution

Step 1: Write the two half-reactions for the voltaic cell:

Ag+(aq) + e−→Ag(s) E◦= 0.80 V

Cu2+(aq) + 2e−→Cu(s) E◦= 0.34 V

Step 2: Calculate the cell potential, Ecell, using the Nernst equation:

Ecell =E◦−0.0592

nlog [Ag+]

[Cu2+]2

Step 3: Determine the number of electrons transferred in the balanced re-

action. In this case, 2 moles of electrons are transferred in the reduction of 1

mole of Cu2+ ions.

Step 4: Calculate the cell potential at 25◦C:

Ecell = (0.80 V) −0.0592

2log 1.0 M

(1.0 M)2

= 0.80 V −0.0296 log(1.0)

= 0.80 V −0.0296(0)

= 0.80 V

Therefore, the cell potential at 25◦C is 0.80 V.

Question 23

Question

Calculate the cell potential for the following reaction at 25

2 AgCl(s)→2 Ag(s) + Cl2(g)

Given:

E◦

cell = 0.16 V

n= 2

Q= 1.0×10−5

Solution

Step 1: Write the Nernst equation

Ecell =E◦

cell −0.0592

nlog Q

Step 2: Substitute the given values into the Nernst equation

Ecell = 0.16 V −0.0592

2log1.0×10−5

Step 3: Calculate the cell potential

Ecell = 0.16 V −0.0296 log1.0×10−5

Ecell = 0.16 V −0.0296 ×(−5)

Ecell = 0.16 V + 0.148

Ecell = 0.308 V

Therefore, the cell potential for the given reaction at 25

C is 0.308 V.

Question 24

Question

A galvanic cell consists of a standard hydrogen electrode (SHE) and a copper

electrode. The half-reaction occurring at the copper electrode is:

Cu2+ + 2e−→Cu(s)

The initial concentration of Cu2+ is 0.0100 M. At 25

C, the cell potential

is measured to be 0.66 V. Determine the concentration of Cu2+ when the cell

potential becomes 0.62 V.

Given: E◦

cell = 0.34 V R= 8.314 J/(mol·K) F= 96,485 C/mol

Solution

Step 1: Begin by writing the Nernst equation, which relates the cell potential

to the concentrations of species involved in the redox reaction:

Ecell =E◦

cell −RT

nF ln [Cu2+]

[Cu] 

Step 2: Substitute the known values into the equation and solve for [Cu2+]:

0.62V= 0.34V−(8.314 J/(mol ·K)) ·(298K)

2·(96485 C/mol) ln x

0.0100

Step 3: Simplify the equation:

0.62V= 0.34V−2473 J/mol

1932 C/mol ln x

0.0100

Step 4: Rearrange the equation and solve for x:

ln x

0.0100=(0.34 −0.62) V ·1932 C/mol

−2473 J/mol

Step 5: Calculate the value of x:

x= 0.0100 ×e





(0.34 −0.62) V ·1932 C/mol

−2473 J/mol





Step 6: Determine the concentration of Cu2+:

x= 0.0100 ×e





(0.34 −0.62) V ·1932 C/mol

−2473 J/mol



≈0.0036 M

Therefore, the concentration of Cu2+ when the cell potential becomes 0.62

V is approximately 0.0036 M.

Question 25

Question

A concentration cell is set up with two half-cells, each containing a Mn3+/Mn2+

half-reaction. One half-cell has a Mn3+ concentration of 0.10 M and the other

half-cell has a Mn3+ concentration of 0.0010 M. Calculate the cell potential

at 25

C for this concentration cell. The standard reduction potential for the

Mn3+/Mn2+ half-reaction is +1.57 V.

Solution

Step 1: Write the balanced half-reaction for the Mn3+/Mn2+ system. The

balanced half-reaction is:

Mn3+ + 1e−→Mn2+

Step 2: Calculate the cell potential using the Nernst equation: The Nernst

equation is given by:

E=E◦−RT

nF ln(Q)

Where: - E= cell potential - E◦= standard cell potential - R= gas constant

(8.314 J/(mol

K)) - T= temperature in Kelvin (25

C = 298 K) - n= number

of moles of electrons transferred in the balanced equation (1 in this case) - F=

Faraday’s constant (96485 C/mol) - Q= reaction quotient

Step 3: Calculate the reaction quotient Qfor the cell.

Q=[Mn2+]high

[Mn2+]low

Q=0.10

0.0010 = 100

Step 4: Substitute the given values into the Nernst equation and solve for

E= 1.57 V −(8.314 J/(mol

K))(298 K)

(1)(96485 C/mol) ln(100)

E= 1.57 V −2476.572

96485 ln(100) V

E≈1.27 V

Therefore, the cell potential at 25

C for this concentration cell is approxi-

mately 1.27 V.

Question 26

Question

Calculate the standard cell potential for the following reaction:

2 Fe3+(aq) + 2 I−(aq)→2 Fe2+(aq)+I2(s)

Given: E◦

Fe3+/Fe2+ = 0.77 V, [Fe3+]=0.010 M, [I−]=0.20 M and the standard

reduction potential for I2/2 I−is 0.54 V.

Solution

Step 1: Write the half-reactions and determine the cell potential.

Cathode: 2 Fe3+(aq) + 2 e−→2 Fe2+(aq)E◦

cathode = 0.77 V

Anode: 2 I−(aq)→I2(s) + 2 e−E◦

anode =−0.54 V

The cell potential is given by E◦

cell =E◦

cathode −E◦

anode. Substitute the given

values:

E◦

cell = 0.77 V −(−0.54 V) = 1.31 V

Step 2: Calculate the reaction quotient (Q) for the cell reaction. The reaction

quotient is given by:

Q=[Fe2+]2[I2]

[Fe3+]2[I−]2

Substitute the given concentrations:

Q=(0.010 M)2(1)

(0.20 M)2= 0.0025

Step 3: Use the Nernst equation to find the cell potential at non-standard

conditions. The Nernst equation is given by:

E=E◦−0.0592

nlog Q

where nis the number of moles of electrons transferred in the balanced redox

reaction. For the given cell reaction, n= 2. Substitute the values into the

Nernst equation:

E= 1.31 V −0.0592

2log 0.0025

E= 1.31 V −0.0298 log 0.0025

E≈1.31 V + 0.0872

E≈1.40 V

Therefore, the standard cell potential for the given reaction is 1.31 V and

the cell potential under the specified non-standard conditions is approximately

1.40 V.

Question 27

Question

A concentration cell is set up with two half-cells: one containing a zinc electrode

in a 0.10 M Zn(NO3)2solution and the other containing a zinc electrode in a

0.0010 M Zn(NO3)2solution. Calculate the cell potential for this concentration

cell at 25

C. The standard reduction potential for the Zn2+/Zn half-cell is -0.76

Solution

Step 1: Write the half-reactions for the oxidation and reduction half-cells. The

overall reaction for the cell involves the oxidation of Zn in the 0.10 M solution

and the reduction of Zn2+ ions in the 0.0010 M solution: Oxidation: Zn(s) →

Zn2+(aq) + 2e−Reduction: Zn2+(aq) + 2e−→Zn(s)

Step 2: Calculate the standard cell potential using the Nernst equation: E◦

= E◦

red - E◦

ox. E◦= 0.00 V - (-0.76 V) = 0.76 V

Step 3: Calculate the cell potential using the Nernst equation: E = E◦-

0.0592

nlog [Zn2+]low

[Zn2+]high . Here, n = 2 (from the balanced half-reactions), [Zn2+]low

= 0.0010 M, and [Zn2+]high = 0.10 M.

Step 4: Calculate the cell potential. E = 0.76 V - 0.0592

2log 0.0010

0.10 E =

0.76 V - 0.0296 log(0.01) E = 0.76 V - 0.0296 ×(-2) E = 0.76 V + 0.0592 E =

0.82 V

Therefore, the cell potential for this concentration cell at 25

C is 0.82 V.

Question 28

Question

A concentration cell is set up using two silver-silver chloride electrodes, one with

0.10 M AgCl

and the other with

1.0×10−3M AgCl

. If the standard reduction potential for the silver-silver chloride electrode is

+0.222 V

25◦C

, calculate the cell potential at

25◦C

Solution

Step 1: Write the overall cell reaction for the concentration cell. Label the

anode and cathode. The overall cell reaction for the concentration cell is:

AgCl(s) →Ag+(aq) + Cl−(aq)

In the cell with

0.10 M AgCl

acting as the anode and

1.0×10−3M AgCl

acting as the cathode, the anode is the electrode with higher concentration and

the cathode is the electrode with lower concentration.

Step 2: Write half-reactions for the anode and cathode, including the stan-

dard reduction potential. The half-reaction for the anode is:

AgCl(s) + e−→Ag+(aq) + Cl−(aq)

The half-reaction for the cathode is:

Ag+(aq)+e−→Ag(s)

Step 3: Calculate the cell potential at

25◦C

using the Nernst equation: The Nernst equation is given by:

E=E◦−0.0592

nlog [oxidized form]

[reduced form] 

For the anode half-reaction:

E◦

anode = +0.222 V

For the cathode half-reaction:

E◦

cathode = +0.222 V

The number of electrons transferred in each half-reaction (n) is 1.

Plugging the values into the Nernst equation for Ecell:

Ecell =Ecathode −Eanode =E◦

cathode −E◦

anode −0.0592

1log 1.0×10−3

0.10 

Ecell = 0.222 V −0.222 V −0.0592 log 1.0×10−3

0.10 

Ecell =−0.0592 log(0.01)

Ecell =−0.0592 × −2

Ecell = 0.1184 V

Therefore, the cell potential at

25◦C

for the concentration cell is

0.1184 V

Question 29

Question

A voltaic cell consists of a standard hydrogen electrode (SHE) in one half-cell

and a copper metal electrode in the other half-cell. The initial concentrations

are [Cu2+]=0.20 M and [H+]=1.0×10−3M. If the cell potential is measured

to be 0.45 V at 25◦C, calculate the standard reduction potential for the copper

half-reaction (Cu2+ + 2e−→Cu).

Solution

Step 1: Write the balanced equation for the overall reaction of the cell and the

half-reactions for the copper electrode: The overall cell reaction is:

2H++ Cu2+ →H2+ Cu

The half-reactions are: Cathode: Cu2+ + 2e−→Cu

Anode: 2H++ 2e−→H2

Step 2: Calculate the standard cell potential (E◦

cell) using the given reduction

potentials for the standard hydrogen electrode (E◦

SHE = 0 V) and the half-

reaction for the copper electrode. The standard cell potential is given by the

Nernst equation:

E◦

cell =E◦

cathode −E◦

anode

where E◦

cathode and E◦

anode are the standard reduction potentials for the cathode

and anode, respectively.

Given that E◦

cell = 0.45 V and E◦

SHE = 0 V, we can rewrite the Nernst

equation as:

0.45 V = E◦

Cu −0

E◦

Cu = 0.45 V

Therefore, the standard reduction potential for the copper half-reaction is

0.45 V.

Question 30

Question

A galvanic cell is constructed with silver metal and a silver ion solution, as well

as copper metal and a copper ion solution. The standard reduction potential

for the silver ion solution is E◦= 0.80 V and for the copper ion solution is

E◦= 0.34 V. The initial concentrations are [Ag+] = 1.0 M, [Cu2+] = 0.10 M.

Calculate the cell potential at 25

C after the cell has run for 15 minutes if

both solutions are at 25

C. Assume the exchange of ions is not limited by any

resistance.

Solution

Step 1: Write the overall cell reaction for the galvanic cell. The overall cell

reaction is the reduction of the silver ion solution and the oxidation of the

copper metal:

Ag+(aq)+e−−→ Ag(s)

Cu(s)−→ Cu2+(aq) + 2e−

Step 2: Calculate the cell potential at standard conditions. The standard

cell potential is given by the equation:

E◦

cell =E◦

cathode −E◦

anode

Substitute the given values:

E◦

cell = 0.80 V −0.34 V = 0.46 V

Step 3: Calculate the reaction quotient, Q. The reaction quotient Q at any

time t is given by:

Q=[Ag](t)

[Cu2+](t)2

Given that [Ag](t)=0.90 M and [Cu2+](t)=0.07 M, we can find Q.

Step 4: Calculate the cell potential at non-standard conditions using the

Nernst equation. The Nernst equation relates the standard cell potential to the

cell potential under non-standard conditions:

Ecell =E◦

cell −0.0592

nlog Q

K

Substitute the values of E◦

cell, Q, n, and K, and calculate Ecell.

Step 5: Calculate the cell potential after running for 15 minutes. Given

that the cell has run for 15 minutes, we can assume the system has reached

equilibrium and calculate the cell potential using Ecell obtained in the previous

step.

Question 31

Question

Calculate the cell potential (Ecell) for the following reaction at 25

Cu2+(0.010M) + Fe(s)→Cu(s) + Fe2+(1.0M)

Given the standard reduction potentials: E◦

Cu2+/Cu(s)= 0.34 V and E◦

Fe2+/Fe(s)=

−0.44 V.

Solution

Step 1: Write the half-reactions and their standard reduction potentials. The

half-reactions for the given reaction are:

Cu2+ + 2e−→Cu E◦

Cu2+/Cu(s)= 0.34 V

Fe2+ + 2e−→Fe E◦

Fe2+/Fe(s)=−0.44 V

Step 2: Calculate the cell potential using the Nernst equation:

The Nernst equation is given by:

Ecell =E◦

cell −0.0592

nlog Q

where Ecell is the cell potential, E◦

cell is the standard cell potential, nis the

number of electrons transferred in the balanced equation, and Qis the reaction

quotient.

Step 3: Calculate the standard cell potential (E◦

cell).

The standard cell potential is the difference between the standard reduction

potentials of the two half-reactions:

E◦

cell =E◦

cathode −E◦

anode

Plugging in the values:

E◦

cell = 0.34 V −(−0.44 V) = 0.78 V

Step 4: Calculate the reaction quotient (Q).

The reaction quotient is given by the concentration of products raised to

their stoichiometric coefficients divided by the concentration of reactants raised

to their stoichiometric coefficients:

Q=[Cu]α[Fe2+]β

[Cu2+]γ

In this case, α= 1, β= 1, and γ= 1.

Plugging in the concentrations:

Q=(1.0)(1.0)

(0.010) = 100

Step 5: Calculate the cell potential (Ecell).

Ecell = 0.78 V −0.0592

2log 100

Ecell = 0.78 V −0.0296 log 100

Ecell = 0.78 V −0.0296(2)

Ecell = 0.78 V −0.0592

Ecell = 0.7208 V

Therefore, the cell potential Ecell for the given reaction at 25

C is 0.7208 V.

Question 32

Question

A half-cell is set up with a silver electrode and a Ag+ion in a 0.50 M solution.

The half-cell is connected to a standard hydrogen electrode (Pt|H2(g, 1 atm)|H+(1 M))

through a salt bridge. Given that E0

Ag+/Ag = 0.80 V and the temperature is

25◦C, calculate the cell potential, Ecell, under standard conditions.

Solution

Step 1: Write the half-reactions for each electrode.

The half-reaction for the silver electrode is:

Ag++ e−→Ag

The half-reaction for the standard hydrogen electrode is:

H++ e−→1

2H2

Step 2: Write the cell reaction.

The cell reaction is the sum of the two half-reactions:

Ag++ H+→Ag + 1

2H2

Step 3: Calculate the cell potential under standard conditions using the

Nernst equation:

Ecell =E0

cell −0.0592

nlog [Ag][H+]

[Ag+][H2]1/2

Since the cell is set up with Ag and Ag+, the concentrations of Ag and Ag+

are equal in the equation. Therefore, the Nernst term simplifies to:

[H+]

[H2]1/2

Given that [H+] = 1 M for the standard hydrogen electrode, the Nernst term

becomes: 1

(1)1/2= 1

Thus, the cell potential under standard conditions is simply the standard

cell potential:

Ecell =E0

cell =E0

Ag+/Ag = 0.80 V

Therefore, the cell potential under standard conditions is 0.80 V.

Question 33

Question

A concentration cell is set up with two half-cells. In one half-cell, the concentra-

tion of Mn2+ is 0.10 M, and in the other half-cell, the concentration of Mn2+ is

1.0×10−3M. If E

for the reaction is E=−1.18 V, calculate the cell potential

at 298 K.

Solution

Step 1: Given the concentrations of Mn2+ ions and E

for the reaction, we can

use the Nernst equation to calculate the cell potential at 298 K. The Nernst

equation is given by:

E=E−0.0592

nlog [Mn2+]right

[Mn2+]left 

Where: - Eis the cell potential. - Eis the standard cell potential. - nis the

number of moles of electrons transferred in the balanced chemical equation. -

[Mn2+]right is the concentration of Mn2+ on the right half-cell. - [Mn2+]left is

the concentration of Mn2+ on the left half-cell.

Step 2: First, let’s calculate the number of moles of electrons transferred in

the balanced chemical equation. Since we don’t have the balanced equation, it

is generally equal to the absolute value of the stoichiometric coefficient of e−.

So n= 1 in this case.

Step 3: Plug in the values into the Nernst equation and solve for the cell

potential E:

E=−1.18 −0.0592

1log 1.0×10−3

0.10 

Step 4: Calculate the logarithmic term:

log 1.0×10−3

0.10 = log1.0×10−2=−2

Step 5: Substitute the calculated logarithmic term back into the Nernst

equation:

E=−1.18 −0.0592(−2)

Step 6: Calculate the cell potential E:

E=−1.18 + 0.1184 = −1.0616 V

Therefore, the cell potential at 298 K is −1.0616 V.

Question 34

Question

Calculate the cell potential at 25

C for a galvanic cell where the following half-

reactions are involved:

Zn2+ + 2e−→Zn(s) E◦=−0.76 V

Fe3+ +e−→Fe2+ E◦= 0.77 V

Given that the concentration of Zn2+ is 2.0×10−2M and the concentration of

Fe3+ is 1.0×10−4M.

Solution

Step 1: Write the overall cell reaction by adding the two half-reactions. The Fe

half-reaction needs to be multiplied by 2 to balance the electrons:

Zn2+ + Fe3+ →Zn(s) + Fe2+

Step 2: Determine the standard cell potential E◦

cell using the given standard

reduction potentials: E◦

cell =E◦

cathode −E◦

anode

E◦

cell = 0.77 −(−0.76)

E◦

cell = 1.53 V

Step 3: Calculate the reaction quotient Qusing the concentrations provided:

Q=[Fe2+]

[Zn2+][Fe3+]=1.0×10−4

(2.0×10−2)(1.0×10−4)= 0.5

Step 4: Use the Nernst equation to calculate the cell potential Ecell at 25

Ecell =E◦

cell −0.0592

nlog Q

Where nis the number of electrons transferred in the balanced cell reaction.

Step 5: Since 2 moles of electrons are transferred in this reaction, substitute

n= 2 and Q= 0.5 into the Nernst equation:

Ecell = 1.53 −0.0592

2log(0.5)

Ecell = 1.53 −0.0296 ×(−0.30)

Ecell = 1.53 + 0.0089

Ecell = 1.5389 V

Therefore, the cell potential at 25

C for this galvanic cell is 1.5389 V.

Question 35

Question

A concentration cell is set up with two half-cells, one containing a 0.10 M solu-

tion of Fe2+ and the other containing a 0.0010 M solution of Fe2+. Calculate

the cell potential at 25

C for this concentration cell. Given that the stan-

dard reduction potential for Fe2+/Fe is +0.77 V and that the gas constant

R= 8.314 J/mol

Solution

Step 1: Write the half-reactions for the cell and find the overall cell reaction.

The half-reactions for the cell are:

Fe2+(0.10 M) + 2e−→Fe(s)E◦

cathode = +0.77 V

Fe2+(0.0010 M) + 2e−→Fe(s)E◦

anode = +0.77 V

The overall cell reaction is the combination of these two half-reactions:

Fe2+(0.10 M) →Fe2+(0.0010 M)

Step 2: Calculate the cell potential using the Nernst equation. The Nernst

equation relates the cell potential to the standard cell potential and the concen-

trations of the reactants in a cell:

E=E◦−0.0592

nlog [Fe2+]anode

[Fe2+]cathode 

Where: - Eis the cell potential, - E◦is the standard cell potential, - nis

the number of moles of electrons transferred in the balanced equation (in this

case, n= 2), - [Fe2+]anode is the concentration of Fe2+ at the anode, and -

[Fe2+]cathode is the concentration of Fe2+ at the cathode.

Plugging in the values:

E= +0.77 V −0.0592

2log 0.0010

0.10 

E= +0.77 V + 0.0296 log(0.01)

E= +0.77 V + 0.0296(−2)

E= +0.77 V −0.0592

E= +0.7108 V

Therefore, the cell potential at 25

C for this concentration cell is +0.7108 V .

CHEM 107 - ESSENTIALS OF GENERAL AND ORGANIC CHEMISTRY Nernst equation Question Bank Set 1

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