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Single electrode potential; Ni/Ag cell EMF

Short answer

Single electrode potential originates from the equilibrium of metal dissolution and ion deposition forming a double layer. Ni has higher oxidation potential and is anode. Cell: Ni | Ni2+ (0.8 M) || Ag+ (0.2 M) | Ag. EMF at 27°C = 1.05 – (0.01292) ln(20) = 1.01 V.

How is single electrode potential originate? From the given electrode system, answer the following questions. Ni → Ni²⁺ + 2e⁻ E° = +0.25 V Ag → Ag⁺ + e⁻ E° = −0.80 V [Ni²⁺] = 0.8 M, [Ag⁺] = 0.2 M, R = 8.314 JK⁻¹mol⁻¹ a) Which of the given electrodes acts as anode and why? b) Construct a galvanic cell using proper cell notation when the given electrodes are coupled. c) Calculate the EMF of the cell at 27°C.

Origin of single electrode potential

When a metal M is dipped in a solution of its ions Mⁿ⁺, two processes occur: M → Mⁿ⁺ + ne⁻ (metal becomes negatively charged) and Mⁿ⁺ + ne⁻ → M (metal becomes positively charged). At equilibrium an electrical double layer forms at the interface, and the potential difference across it is the single electrode potential.

The values given are oxidation potentials: Ni +0.25 V, Ag −0.80 V.

a) Anode

Ni has the higher oxidation potential (+0.25 V > −0.80 V), so it is oxidised more easily. Ni is the anode; Ag is the cathode.

b) Cell notation

Ni | Ni²⁺ (0.8 M) || Ag⁺ (0.2 M) | Ag

Anode   : Ni → Ni²⁺ + 2e⁻
Cathode : 2Ag⁺ + 2e⁻ → 2Ag
Cell    : Ni + 2Ag⁺ → Ni²⁺ + 2Ag    (n = 2)

c) EMF at 27°C (300 K)

E°cell = 0.25 + 0.80 = 1.05 V
Q = [Ni²⁺]/[Ag⁺]² = 0.8/(0.2)² = 20
RT/nF = 8.314 × 300/(2 × 96500) = 0.01292 V
E = E° − (RT/nF) ln Q = 1.05 − 0.01292 × 2.996
  = 1.05 − 0.0387 = 1.011 V

Answer: EMF ≈ 1.01 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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