Solved Examples (Dedicated Problem Set)

How to Use This Problem Set

This is your full workout for Electrochemistry, grouped by theme: cell potentials and feasibility, Nernst-equation calculations, ΔG–Kc links, conductance and molar conductivity, Kohlrausch's law, and Faraday's-law electrolysis.

Solve each with a pen before reading the worked solution. Keep these constants handy:

• $F = 96500$ C mol⁻¹; $\dfrac{2.303RT}{F}=0.059$ V at 298 K.
• $\Lambda_m = \dfrac{\kappa\times1000}{c}$ ($\kappa$ in S cm⁻¹, c in mol L⁻¹).
• $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$; $\Delta_r G^\circ=-nFE^\circ_{cell}$; $E^\circ_{cell}=\dfrac{0.059}{n}\log K_c$.

Standard reduction potentials used below (V): Zn²⁺/Zn −0.76, Fe²⁺/Fe −0.44, Ni²⁺/Ni −0.25, Cu²⁺/Cu +0.34, Ag⁺/Ag +0.80, Mg²⁺/Mg −2.36.

Solved Examples — Cell Potential & Feasibility

Example 1. Find $E^\circ_{cell}$ for Zn | Zn²⁺ || Ag⁺ | Ag.

Solution: Cathode Ag (+0.80), anode Zn (−0.76). $E^\circ_{cell}=0.80-(-0.76)=1.56$ V.

Example 2. Is the reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag feasible?

Solution: $E^\circ_{cell}=0.80-0.34=+0.46$ V (positive) → feasible.

Example 3. Find $E^\circ_{cell}$ for Ni | Ni²⁺ || Cu²⁺ | Cu and state if spontaneous.

Solution: $E^\circ_{cell}=0.34-(-0.25)=+0.59$ V → positive → spontaneous.

Solved Examples — Cell Potential (continued)

Example 4. Will Fe displace Cu from CuSO₄? (Fe²⁺/Fe −0.44, Cu²⁺/Cu +0.34)

Solution: Fe + Cu²⁺ → Fe²⁺ + Cu, $E^\circ_{cell}=0.34-(-0.44)=+0.78$ V → positive → yes.

Example 5. Calculate $E^\circ_{cell}$ for Mg | Mg²⁺ || Cu²⁺ | Cu.

Solution: $E^\circ_{cell}=0.34-(-2.36)=+2.70$ V.

Example 6. Which is the stronger reducing agent, Mg or Zn?

Solution: More negative $E^\circ$ → stronger reducing agent. Mg (−2.36) < Zn (−0.76), so Mg is stronger.

Solved Examples — Nernst Equation

Example 7. For Zn | Zn²⁺(0.01 M) || Cu²⁺(1 M) | Cu, find $E_{cell}$ ($E^\circ=1.10$ V).

Solution: $E=1.10-\dfrac{0.059}{2}\log\dfrac{0.01}{1}=1.10-0.0295(-2)=1.10+0.059=1.159$ V.

Example 8. Potential of a Cu²⁺/Cu electrode in 0.01 M Cu²⁺ ($E^\circ=+0.34$ V).

Solution: $E=0.34-\dfrac{0.059}{2}\log\dfrac{1}{0.01}=0.34-0.0295(2)=0.34-0.059=0.281$ V.

Example 9. Hydrogen electrode potential at pH = 3 (H₂ at 1 bar).

Solution: $E=-0.059\times\text{pH}=-0.059\times3=-0.177$ V.

Solved Examples — Nernst (continued)

Example 10. For Ag | Ag⁺(0.01 M) || Ag⁺(1 M) | Ag (concentration cell, n=1), find $E_{cell}$.

Solution: $E=-0.059\log\dfrac{0.01}{1}=-0.059(-2)=0.118$ V.

Example 11. For Zn | Zn²⁺(1 M) || Cu²⁺(0.001 M) | Cu, find $E_{cell}$ ($E^\circ=1.10$ V).

Solution: $E=1.10-\dfrac{0.059}{2}\log\dfrac{1}{0.001}=1.10-0.0295(3)=1.10-0.0885=1.0115$ V.

Example 12. At what $\log Q$ does the Daniell cell ($E^\circ=1.10$ V, n=2) read $E_{cell}=1.0$ V?

Solution: $1.0=1.10-0.0295\log Q\Rightarrow\log Q=\dfrac{0.10}{0.0295}=3.39$.

Solved Examples — ΔG and Kc

Example 13. $\Delta_r G^\circ$ for a cell with $E^\circ=1.56$ V, n=2 ($F=96500$).

Solution: $\Delta_r G^\circ=-2\times96500\times1.56=-301{,}080$ J $=-301.1$ kJ.

Example 14. $\log K_c$ for a cell with $E^\circ=0.59$ V, n=2.

Solution: $\log K_c=\dfrac{nE^\circ}{0.059}=\dfrac{2\times0.59}{0.059}=20$. So $K_c=10^{20}$.

Example 15. $E^\circ_{cell}$ for a 2-electron cell with $K_c=10^6$.

Solution: $E^\circ=\dfrac{0.059}{2}\log10^6=\dfrac{0.059\times6}{2}=0.177$ V.

Solved Examples — Conductance

Example 16. Conductivity 0.0248 S cm⁻¹, c = 0.2 M. Find $\Lambda_m$.

Solution: $\Lambda_m=\dfrac{0.0248\times1000}{0.2}=124$ S cm² mol⁻¹.

Example 17. A cell of cell constant 1.5 cm⁻¹ has resistance 50 Ω. Find κ.

Solution: $\kappa=G^*/R=1.5/50=0.03$ S cm⁻¹.

Example 18. Conductance of a 0.1 M solution is $5\times10^{-3}$ S in a cell of constant 0.8 cm⁻¹. Find $\Lambda_m$.

Solution: $\kappa=G\times G^*=5\times10^{-3}\times0.8=4\times10^{-3}$ S cm⁻¹. $\Lambda_m=\dfrac{4\times10^{-3}\times1000}{0.1}=40$ S cm² mol⁻¹.

Solved Examples — Kohlrausch's Law

Example 19. $\Lambda^\circ_m$ of CaCl₂ from $\lambda^\circ(\text{Ca}^{2+})=119.0$ and $\lambda^\circ(\text{Cl}^-)=76.3$ S cm² mol⁻¹.

Solution: $\Lambda^\circ_m=119.0+2(76.3)=271.6$ S cm² mol⁻¹.

Example 20. $\Lambda^\circ_m$ of acetic acid from CH₃COONa (91.0) + HCl (425.9) − NaCl (126.4).

Solution: $91.0+425.9-126.4=390.5$ S cm² mol⁻¹.

Example 21. Degree of dissociation when $\Lambda_m=39.05$ and $\Lambda^\circ_m=390.5$ S cm² mol⁻¹.

Solution: $\alpha=\Lambda_m/\Lambda^\circ_m=39.05/390.5=0.10$ (10%).

Solved Examples — Faraday's Laws

Example 22. Charge to deposit 0.5 mol Cu (Cu²⁺ + 2e⁻ → Cu).

Solution: 0.5 mol Cu needs $0.5\times2=1$ mol electrons = $96500$ C.

Example 23. Mass of Ag (M=108) deposited by 9650 C (Ag⁺ + e⁻ → Ag).

Solution: mol e⁻ $=9650/96500=0.1$; mol Ag $=0.1$; mass $=0.1\times108=10.8$ g.

Example 24. A 1.5 A current for 30 min deposits how much Cu (M=63.5)?

Solution: $Q=1.5\times1800=2700$ C; mol e⁻ $=2700/96500=0.02798$; mol Cu $=0.01399$; mass $=0.01399\times63.5=0.888$ g.

Solved Examples — Faraday's Laws (continued)

Example 25. Time for 2 A to deposit 0.50 g Cu (M=63.5, n=2).

Solution: mol Cu $=0.50/63.5=7.874\times10^{-3}$; mol e⁻ $=2\times7.874\times10^{-3}=0.01575$; $Q=0.01575\times96500=1519.7$ C; $t=Q/I=1519.7/2=759.8$ s.

Example 26. Moles of electrons in 4825 C.

Solution: mol e⁻ $=4825/96500=0.05$ mol.

Example 27. Mass of Al (M=27) deposited by 3 F of charge.

Solution: Al³⁺ + 3e⁻ → Al. 3 F = 3 mol electrons → 1 mol Al → mass $=27$ g.

Solved Examples — Mixed

Example 28. Faraday's second law: equal charge deposits 0.36 g Ag (eq. mass 108) and how much Zn (eq. mass 32.7)?

Solution: $m_{Zn}=m_{Ag}\times\dfrac{32.7}{108}=0.36\times0.3028=0.109$ g.

Example 29. $E^\circ_{cell}=0.0295$ V, n=2. Find $K_c$.

Solution: $\log K_c=\dfrac{2\times0.0295}{0.059}=1$; $K_c=10$.

Example 30. A concentration cell Cu|Cu²⁺(0.001 M)||Cu²⁺(0.1 M)|Cu. Find $E_{cell}$.

Solution: $E=-\dfrac{0.059}{2}\log\dfrac{0.001}{0.1}=-0.0295(-2)=0.059$ V.

Solved Examples — Mixed (continued)

Example 31. $\Delta_r G^\circ$ for Zn + Cu²⁺ → Zn²⁺ + Cu ($E^\circ=1.10$ V, n=2, F=96500).

Solution: $\Delta_r G^\circ=-2\times96500\times1.10=-212{,}300$ J $=-212.3$ kJ.

Example 32. Molar conductivity of 0.001 M HCl is 421 S cm² mol⁻¹; its $\Lambda^\circ_m$ is 426 S cm² mol⁻¹. Degree of dissociation?

Solution: $\alpha=421/426=0.988$ (HCl is a strong acid, about 98.8% dissociated at this concentration).

Example 33. Number of electrons (n) in 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O.

Solution: As written, n = 10 electrons are transferred.