---
title: "Ni/Cd galvanic cell: electrodes, emf"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:ni-cd-cell"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ni-cd-cell
site: BCT Engineering
---

# Ni/Cd galvanic cell: electrodes, emf

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2072 Magh, 4 marks
- Video: [Redox potentials, Cell Diagrams and EMF, Cadmium and nickel](https://www.youtube.com/watch?v=nizjgu_5VrE) — jamesmungall
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ni-cd-cell
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

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.

## 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.

## Answer

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
```text
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)
```text
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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Ni oxidation +0.25 V and Cd oxidation +0.40 V. Which is the anode and why?
   A. Ni, because it has the lower oxidation potential
   B. Ni, because it is heavier
   C. Cd, because it has the higher reduction potential
   D. Cd, because it has the higher oxidation potential and loses electrons more easily
   Answer: D. The higher oxidation potential means easier oxidation, so Cd is the anode and Ni the cathode. (Key point: Choosing anode)

2. What is the cell reaction?
   A. Ni + Cd2+ -> Ni2+ + Cd
   B. Cd + Ni -> Cd2+ + Ni2+
   C. Cd2+ + Ni2+ -> Cd + Ni
   D. Cd + Ni2+ -> Cd2+ + Ni
   Answer: D. Cd is oxidised at the anode and Ni2+ is reduced at the cathode. (Key point: Cell reaction)

3. Correct cell notation is:
   A. Ni | Ni2+ (1 M) || Cd2+ (1 M) | Cd
   B. Cd2+ | Cd || Ni | Ni2+
   C. Cd | Ni2+ || Cd2+ | Ni
   D. Cd | Cd2+ (1 M) || Ni2+ (1 M) | Ni
   Answer: D. Anode (oxidation) on the left, cathode on the right. (Key point: Cell notation)

4. At 27°C with 1 M solutions, what is the emf?
   A. 0.65 V
   B. -0.15 V
   C. 0.15 V plus 0.0129 V
   D. 0.15 V
   Answer: D. E°cell = 0.40 + (-0.25) = 0.15 V; Q = 1 so ln Q = 0 and E = E° = 0.15 V. (Key point: EMF calculation)

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