---
title: "Electrolytic cell; [Ni²⁺] from emf of Ni/Cu cell"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:ni-cu-cell-concentration"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ni-cu-cell-concentration
site: BCT Engineering
---

# Electrolytic cell; [Ni²⁺] from emf of Ni/Cu cell

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2069 Poush, 2+3 marks
- Video: [Write the Nernst equation and calculate e.m.f of the following cell of 298 K.Ni|Ni²+ (0.01M)||Cu²+](https://www.youtube.com/watch?v=EYFGA7ux1a8) — Studypur
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ni-cu-cell-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is meant by electrolytic cell and electrode potential? Calculate the concentration of Ni²⁺ in the cell at 25°C, if the emf is 0.601 V.
Ni/Ni²⁺(α=?) // Cu²⁺(0.75M)/Cu
Given, E° Ni/Ni²⁺ = 0.25 V and E° Cu²⁺/Cu = 0.34 V

## Short answer

An electrolytic cell uses electrical energy to drive a non-spontaneous reaction. Electrode potential is the interfacial potential difference. With E°cell = 0.59 V and Nernst equation for Ni + Cu²⁺ → Ni²⁺ + Cu, when EMF = 0.601 V, [Ni²⁺] = 0.32 M.

## Answer

#### Electrolytic cell
A cell in which electrical energy from an external source is used to bring about a non-spontaneous chemical reaction (electrolysis). Here electrical energy is converted into chemical energy; e.g. electrolysis of molten NaCl or of CuSO₄ solution. (Anode is positive and cathode is negative.)

#### Electrode potential
The potential difference set up between a metal and the solution of its ions at the interface, due to the tendency to lose or gain electrons. Standard electrode potential is measured at 1 M and 25°C against SHE.

#### Calculation
Cell: Ni | Ni²⁺ (x) || Cu²⁺ (0.75 M) | Cu. The cell reaction is Ni + Cu²⁺ → Ni²⁺ + Cu (n = 2).
```text
E°cell = 0.25 + 0.34 = 0.59 V
E = E° − (0.0591/2) log ([Ni²⁺]/[Cu²⁺])
0.601 = 0.59 − 0.02955 log (x/0.75)
log (x/0.75) = (0.59 − 0.601)/0.02955 = −0.372
x/0.75 = 10⁻⁰·³⁷² = 0.4246
x = 0.318 M
```
**Answer: [Ni²⁺] ≈ 0.32 M.**

## Check yourself

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

1. An electrolytic cell:
   A. Converts electrical energy to chemical energy
   B. Converts chemical energy to electrical energy
   C. Needs no external source
   D. Has a spontaneous reaction
   Answer: A. External electricity drives a non-spontaneous reaction. (Key point: Electrolytic cell)

2. E°cell for Ni + Cu2+ -> Ni2+ + Cu with E°ox Ni = 0.25 V and E°red Cu = 0.34 V:
   A. 0.59 V
   B. 0.09 V
   C. -0.59 V
   D. 0.34 V
   Answer: A. 0.25 + 0.34 = 0.59 V. (Key point: E°cell)

3. Since E = 0.601 V exceeds E° = 0.59 V, [Ni2+]/[Cu2+] must be:
   A. Less than 1
   B. Greater than 1
   C. Exactly 1
   D. Cannot tell
   Answer: A. E > E° means log Q is negative, so Q < 1. (Key point: Sign of the Nernst term)

4. [Ni2+] =
   A. 0.32 M
   B. 0.75 M
   C. 1.77 M
   D. 0.075 M
   Answer: A. x/0.75 = 0.4246, x = 0.318 M. (Key point: Result)

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