---
title: "Galvanic vs electrolytic cell; Cd/Cu cell emf"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:galvanic-electrolytic-cd-cu"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/galvanic-electrolytic-cd-cu
site: BCT Engineering
---

# Galvanic vs electrolytic cell; Cd/Cu cell emf

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2068 Bhadra, 1+4 marks
- Video: [Electrochemical Cells (Galvanic/Voltaic Cell and Electrolytic Cell)](https://www.youtube.com/watch?v=UowPA1z-c9c) — Kawsar Ahmed
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/galvanic-electrolytic-cd-cu
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

How does a galvanic cell differ from an electrolytic cell? Calculate the emf of the following cell at 25°C giving electrode reactions and cell reaction.
Cd(s)|Cd²⁺(0.01M)//Cu²⁺(0.5M)|Cu(s)
E°Cd²⁺/Cd = −0.140 V, E°Cu²⁺/Cu = 0.34 V

## Short answer

Galvanic cells generate electricity from spontaneous reactions; electrolytic cells consume electricity for non-spontaneous ones. The Cd–Cu cell EMF: E°cell = 0.48 V, using Nernst equation E = 0.48 − (0.0591/2) log(0.01/0.5) = 0.53 V.

## Answer

#### Galvanic cell vs electrolytic cell
| Point | Galvanic cell | Electrolytic cell |
|---|---|---|
| Energy change | Chemical → electrical | Electrical → chemical |
| Reaction | Spontaneous | Non-spontaneous (driven by external source) |
| Anode | Negative | Positive |
| Cathode | Positive | Negative |
| Salt bridge | Needed (two half cells) | Not needed (one cell) |
| Example | Daniell cell | Electrolysis of fused NaCl |

#### EMF of Cd | Cd²⁺ (0.01 M) || Cu²⁺ (0.5 M) | Cu
```text
Anode   : Cd → Cd²⁺ + 2e⁻
Cathode : Cu²⁺ + 2e⁻ → Cu
Cell    : Cd + Cu²⁺ → Cd²⁺ + Cu        (n = 2)

E°cell = E°(Cu²⁺/Cu) − E°(Cd²⁺/Cd) = 0.34 − (−0.140) = 0.48 V
Q = [Cd²⁺]/[Cu²⁺] = 0.01/0.5 = 0.02 ,  log Q = −1.699
E = E° − (0.0591/2) log Q
  = 0.48 + 0.02955 × 1.699
  = 0.48 + 0.0502 = 0.530 V
```
**Answer: EMF ≈ 0.53 V** (E°Cd as given in the question).

## Check yourself

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

1. In an electrolytic cell:
   A. Anode is positive, cathode negative
   B. Anode is negative, cathode positive
   C. Both positive
   D. Both negative
   Answer: A. Opposite to the galvanic cell, where anode is negative. (Key point: Polarity of electrodes)

2. Galvanic cell reaction is:
   A. Spontaneous
   B. Non-spontaneous
   C. Driven by external source
   D. Never occurs
   Answer: A. Chemical energy is converted to electrical energy spontaneously. (Key point: Spontaneity)

3. E°cell = 0.34 - (-0.140) =
   A. 0.48 V
   B. 0.20 V
   C. -0.48 V
   D. 0.34 V
   Answer: A. E°cathode - E°anode = 0.48 V. (Key point: E°cell)

4. Q = 0.01/0.5 = 0.02. E =
   A. 0.53 V
   B. 0.43 V
   C. 0.48 V
   D. 0.58 V
   Answer: A. E = 0.48 + 0.02955 x 1.699 = 0.530 V. (Key point: Final EMF)

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