---
title: "Electrochemical series; Ni galvanic cell emf"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:electrochemical-series-ni-cell"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/electrochemical-series-ni-cell
site: BCT Engineering
---

# Electrochemical series; Ni galvanic cell emf

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2068 Magh, 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/electrochemical-series-ni-cell
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is an electrochemical series? With the help of electrochemical series complete the following galvanic cell and calculate the cell emf.
(−) Ni / Ni²⁺(1.5M) // [?]
E°Ni²⁺/Ni = −0.25 V

## Short answer

The electrochemical series ranks electrodes by increasing standard reduction potential. The Ni anode is paired with a Cu || Cu²⁺ (1 M) cathode. Cell EMF = 0.59 − (0.0591/2) log(1.5/1) ≈ 0.585 V.

## Answer

#### Electrochemical series
The arrangement of elements (electrodes) in increasing order of their standard reduction potentials (or decreasing order of oxidation potential) is the **electrochemical series**. Metals at the top (Li, K, Zn …) are strong reducing agents and lose electrons easily; metals at the bottom (Cu, Ag, Au) are weak reducing agents. A metal higher in the series displaces a lower metal from its salt solution.

Part of the series (E° reduction, V): Zn −0.76 < Ni −0.25 < H₂ 0.00 < Cu +0.34 < Ag +0.80.

#### Completing the cell
Ni (E° = −0.25 V) is the anode, so the right-hand electrode must have a higher reduction potential. Choosing copper (E° = +0.34 V), with 1 M Cu²⁺ (concentration not given, so standard assumed):

(−) Ni | Ni²⁺ (1.5 M) || Cu²⁺ (1 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⁺] = 1.5/1 = 1.5 ,  log Q = 0.176
E = 0.59 − 0.02955 × 0.176 = 0.59 − 0.0052 = 0.585 V
```
**Answer: EMF ≈ 0.585 V.**

## Check yourself

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

1. The electrochemical series arranges electrodes in order of:
   A. Increasing standard reduction potential
   B. Atomic mass
   C. Melting point
   D. Density
   Answer: A. Elements are listed by increasing standard reduction potential. (Key point: Definition)

2. A metal higher in the series:
   A. Displaces a lower metal from its salt solution
   B. Is displaced by lower metals
   C. Is inert
   D. Has higher reduction potential than all
   Answer: A. Higher metals are stronger reducing agents and displace lower ones. (Key point: Displacement)

3. With Ni (E° = -0.25 V) as anode, the cathode must have:
   A. Higher reduction potential, e.g. Cu (+0.34 V)
   B. Lower reduction potential, e.g. Zn (-0.76 V)
   C. The same potential
   D. No potential
   Answer: A. The cathode has a higher reduction potential than the anode. (Key point: Completing the cell)

4. Ni | Ni2+ (1.5 M) || Cu2+ (1 M) | Cu: E =
   A. 0.585 V
   B. 0.59 V
   C. 0.60 V
   D. 0.50 V
   Answer: A. 0.59 - 0.02955 x log 1.5 (0.176) = 0.585 V. (Key point: EMF)

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