---
title: "Normal hydrogen electrode; Ni/Cu cell emf"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:nhe-ni-cu-cell"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/nhe-ni-cu-cell
site: BCT Engineering
---

# Normal hydrogen electrode; Ni/Cu cell emf

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2067 Chaitra, 2+3 marks
- Video: [Cell Potential Problems - Electrochemistry](https://www.youtube.com/watch?v=UzkLP8segcs) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/nhe-ni-cu-cell
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is normal hydrogen electrode? A galvanic cell consists of a metallic nickel plate immersed in 0.1M Ni(NO₃)₂ solution and a metallic plate of copper in 0.2M CuSO₄ solution. Calculate the emf of this cell.
E°Ni²⁺/Ni = −0.25 V, E°Cu²⁺/Cu = +0.34 V

## Short answer

Normal hydrogen electrode consists of Pt foil coated with Pt black, immersed in 1 M H⁺ with H₂ at 1 atm and 25°C; potential taken as zero. For Ni|Ni²⁺(0.1 M)||Cu²⁺(0.2 M)|Cu, Ni anode, Cu cathode: E°=0.59 V, Q=0.5, EMF=0.60 V using Nernst equation.

## Answer

#### Normal hydrogen electrode
A reference electrode made of platinum foil coated with platinum black, dipped in 1 N (1 M) H⁺ solution with pure H₂ gas at 1 atm bubbled around it at 25°C. Its potential is taken as zero. Notation: Pt, H₂ (1 atm) | H⁺ (1 M).

#### EMF of the Ni–Cu cell
Ni | Ni²⁺ (0.1 M) || Cu²⁺ (0.2 M) | Cu
```text
Anode   : Ni → Ni²⁺ + 2e⁻
Cathode : Cu²⁺ + 2e⁻ → Cu
Cell    : Ni + Cu²⁺ → Ni²⁺ + Cu         (n = 2)

E°cell = 0.34 − (−0.25) = 0.59 V
Q = [Ni²⁺]/[Cu²⁺] = 0.1/0.2 = 0.5 ,  log Q = −0.301
E = 0.59 − 0.02955 × (−0.301)
  = 0.59 + 0.0089 = 0.599 V
```
**Answer: EMF ≈ 0.60 V.**

## Check yourself

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

1. Normal hydrogen electrode has potential:
   A. Zero
   B. +0.34 V
   C. -0.25 V
   D. 1 V
   Answer: A. It is the reference taken as 0. (Key point: NHE)

2. E°cell for Ni-Cu: 0.34 - (-0.25) =
   A. 0.59 V
   B. 0.09 V
   C. -0.59 V
   D. 0.34 V
   Answer: A. E°cathode - E°anode. (Key point: E°cell)

3. Q = [Ni2+]/[Cu2+] = 0.1/0.2 =
   A. 0.5
   B. 2
   C. 0.02
   D. 5
   Answer: A. Q = 0.5, log Q = -0.301. (Key point: Q)

4. E = 0.59 - 0.02955 x (-0.301) =
   A. 0.60 V
   B. 0.58 V
   C. 0.59 V
   D. 0.56 V
   Answer: A. 0.59 + 0.0089 = 0.599, about 0.60 V. (Key point: EMF)

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