Chemistry Questions and Answers – Electrochemistry – Nernst Equation

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This set of Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Electrochemistry – Nernst Equation”.

1. The standard oxidation potential of Ni/Ni2+ electrode is 0.3 V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution with the measured e.m.f. be zero at 25°C? (Assume [Ni2+] = 1M)
a) 5.08
b) 4
c) 4.5
d) 5.25
View Answer

Answer: a
Explanation: Given,
Standard oxidation potential of Ni/Ni2+ electrode, E°OP = 0.3 V,
Ni → Ni2+ + 2e
2H+ + 2e → H2
cell = E° (OP) + E° (RP)
(cell) = 0.3 + 0.0 = 0.3 V
According to Nernst equation, E (cell) = E°(cell) + \(\frac{0.059}{2}\) log10([H+]2 / [Ni]+2)
0 = 0.3 + \(\frac{0.059}{2}\)log10([H+]2)
-log ([H+]) = 5.08
pH = 5.08.

2. Calculate the equilibrium constant for the reaction Fe + CuSO4 ⇌ FeSO4 + Cu at 25°C.
(Given E°(OP/Fe) = 0.5 V°, E°(OP/Cu) = -0.4 V)
a) 3.46 × 1030
b) 3.46 × 1026
c) 3.22 × 1030
d) 3.22 × 1026
View Answer

Answer: c
Explanation: The cell reaction shows oxidation of Fe and reduction of Cu2+
Therefore for the reaction, Fe + CuSO4 ⇌ FeSO4 + Cu
(cell) = E°(OP/Fe) + E°(RP/Cu)
(cell) = 0.5 + 0.4 = 0.9 V
We have, E° = \(\frac{0.059}{2}\)log10 Kc
0.9 = \(\frac{0.059}{2}\)log10 Kc
Kc = 3.22 × 1030.

3. Calculate the e.m.f. of the half-cell given below.
Pt, H2 | HCl at 1-atmosphere pressure and 0.1 M. Given, E°(OP) = 2 V.
a) 4 V
b) 5.6 V
c) 3.4 V
d) 5.4 V
View Answer

Answer: d
Explanation: Given, E°(OP) = 2 V,
H2 → 2H+ + 2e
E(OP) = E°(OP) – \(\frac{0.059}{2}\)log10([H+]2/ P (H2))
E(OP) = 2 – \(\frac{0.059}{2}\)log10 (0.022 / 1) = 5.4 V.
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4. The equilibrium constant for a cell reaction, Cu(g) + 2Ag+(aq) → Cu2+(aq) + 2Ag (s) is 4 × 1016. Find E° (cell) for the cell reaction.
a) 0.63 V
b) 0.49 V
c) 1.23 V
d) 3.24 V
View Answer

Answer: b
Explanation: Given, equilibrium constant Kc = 4 × 1016
(cell) = \(\frac{0.059}{2}\)log10 Kc
(cell) = \(\frac{0.059}{2}\)log10 (4 × 1016)=0.49V.

5. What is the correct Nernst equation for M2+ (aq) + 2e+ → M (s) at 45°C?
a) E°(M2+/M) + 0.315log10 (1 / [M]+2)
b) E° (M2+/M) + 0.0425log10 (1 / [M]+2)
c) E° (M2+/M) + 0.0315log10 (1 / [M]+2)
d) E° (M2+/M) + 0.0326log10 (1 / [M]+2)
View Answer

Answer: c
Explanation: Given, Temperature T = 45°C
We know, n (number of electrons transferred) = 2
According to Nernst equation, E(M2+/M) = E° (M2+/M) +2.303\(\frac{RT}{nF}\)log10 (M / [M]+2)
Concentration of [M] is taken to be 1
The equation becomes: E° (M2+/M) +2.303 \(\frac{RT}{nF}\)log10 (1/[M]+2)
E(M2+/M) = E° (M2+/M) +2.303 × \(\frac{8.314 \times 318}{2 \times 96500}\)log10 (1 / [M]+2) =E° (M2+/M) + 0.0315log10 (1 / [M]+2).
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6. The e.m.f and the standard e.m.f of a cell in the following reaction is 5 V and 5.06 V at room temperature, Ni(s) + 2Ag+(n) → Ni2+(0.02M) + 2Ag(s). What is the concentration of Ag+ ions?
a) 0.0125 M
b) 0.0314 M
c) 0.0625 M
d) 0.0174 M
View Answer

Answer: d
Explanation: Given, Temperature T = 298K
Concentration of Ni2+ = (0.02M)
E(cell) = E°(cell) – \(\frac{0.059}{n}\)log10 (Anode / Cathode)
5 = 5.06 – \(\frac{0.059}{2}\)log10(0.02 / [Ag+]2)
[Ag+]2 = 0.0174 M.

7. Calculate the electrode potential of the given electrode.
Pt, Cl2(2 bar)| 2Cl(0.02 M); E°(Cl2 | 2Cl) = 3.4 V
a) 3.51 V
b) 3.55 V
c) 1.26 V
d) 2.95 V
View Answer

Answer: a
Explanation: The electrode reaction is Cl2(g) + 2e → 2Cl
E = E˚ – \(\frac{0.059}{n}\)log ([Cl]2 / P(Cl₂))
E = 3.4 – \(\frac{0.059}{2}\)log (0.022 / 2) = 3.51 V.
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8. A zinc rod dipped in n molar solution of ZnSO4 has an electrode potential of -0.56 V. The salt is 98 percent dissociated at room temperature. What is the molarity of the solution? (E°(Zn+2/Zn) = -0.5 V)
a) 8.44 × 10-3 M
b) 9.44 × 10-4 M
c) 8.44 × 10-4 M
d) 9.44 × 10-3 M
View Answer

Answer: d
Explanation: The electrode reaction is Zn2+ + 2e → Zn
The number of electrons transferred, n= 2
Applying Nernst equation, we get E(Zn+2/Zn) = E°(Zn+2/Zn) – \(\frac{0.059}{2}\)log (1 / [Zn2+])
[Zn2+] = \(\frac{98}{100}\) × n = 0.98n M
-0.56 = -0.5 – \(\frac{0.059}{2}\)log (1 / 0.98n)
n= 9.44 × 10-3 M.

9. What is the pH of HCl solution when the hydrogen gas electrode shows a potential of -0.22 V at standard temperature and pressure?
a) 2.17
b) 2.98
c) 3.73
d) 3.14
View Answer

Answer: c
Explanation: Given, potential of hydrogen gas electrode = -0.22 V
Electrode reaction: H+ + e → 0.5 H2
Applying Nernst equation,
E(H+/H₂) = E°(H+/H2) – 0.059 log (1/ [H+])
(H+/H₂) = 0 for hydrogen gas electrode
-0.22 = 0.059 log H+
-0.22 = -0.059pH
pH= 3.73.
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10. What is the value of universal gas constant in Nernst equation when the potential is given in volts?
a) 8.314 J mol-1K-1
b) 0.0821 L atm mol-1K-1
c) 8.205 m3 atm mol-1K-1
d) 1.987 cal mol-1K-1
View Answer

Answer: a
Explanation: The universal gas constant is denoted by R and is expressed in units of energy per temperature per mole. Since volts is the SI-Unit of potential, R must also be taken in SI-Units which is J mol-1K-1.

11. What is the number of electrons transferred in an equation if the Nernst equation is E(cell) = E°(cell) – 9.83 × 10-3 × log10 (Anode / Cathode)?
a) 2
b) 6
c) 4
d) 1
View Answer

Answer: b
Explanation: Nernst equation = E°(cell) – \(\frac{0.059}{n}\) log10(Anode / Cathode)
On comparing both the formulae, \(\frac{0.059}{n}\) = 9.83 × 10-3
n= 6.

12. Find the number of electrons transferred in the equation Cu(g) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s).
a) 4
b) 3
c) 2
d) 1
View Answer

Answer: c
Explanation: 2Ag+(aq) + 2e → 2Ag(s)
From the equation it is evident that 2Ag+ takes 2 electrons from Cu and neutralizes to form 2Ag.

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