Skip to main content

Ni/Cd galvanic cell: electrodes, emf

Short answer

Cd has the higher oxidation potential (+0.40 V) and acts as anode; Ni is cathode. Cell reaction: Cd + Ni²⁺ → Cd²⁺ + Ni. Notation: Cd|Cd²⁺||Ni²⁺|Ni. E°cell = 0.15 V; at 27°C with Q=1, EMF = 0.15 V.

From the given electrode potential values answer the following questions. Ni → Ni²⁺ + 2e⁻ E° = +0.25 V Cd → Cd²⁺ + 2e⁻ E° = +0.40 V Given: R = 8.314 JK⁻¹mol⁻¹, F = 96500 C i) Which of the above electrodes act as anode and which one as cathode and why? ii) Write the electrode and cell reactions when the given electrodes are coupled. iii) Using these electrodes construct a galvanic cell with proper cell notation. iv) Calculate the emf of the cell at 27°C when given electrodes are coupled together.

The values are oxidation potentials: Ni → Ni²⁺ + 2e⁻ (+0.25 V) and Cd → Cd²⁺ + 2e⁻ (+0.40 V). No concentrations are given, so standard conditions (1 M) are assumed.

i) Anode and cathode

Cd has the higher oxidation potential (+0.40 V > +0.25 V), so it loses electrons more easily: Cd is the anode (oxidation). Ni is the cathode, where Ni²⁺ is reduced.

ii) Electrode and cell reactions

Anode   : Cd → Cd²⁺ + 2e⁻
Cathode : Ni²⁺ + 2e⁻ → Ni
Cell    : Cd + Ni²⁺ → Cd²⁺ + Ni

iii) Cell notation

Cd | Cd²⁺ (1 M) || Ni²⁺ (1 M) | Ni

iv) EMF at 27°C (300 K)

E°cell = E°ox(anode) + E°red(cathode)
       = 0.40 + (−0.25) = +0.15 V
E = E° − (RT/nF) ln Q ,  RT/nF = 8.314×300/(2×96500) = 0.01292 V
Q = [Cd²⁺]/[Ni²⁺] = 1/1 = 1  →  ln Q = 0
E = 0.15 V

Answer: EMF = 0.15 V (positive, so the reaction is spontaneous).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Chemistry chapter 1 questions →