JEE Main 2014ChemistryElectrochemistryMediumMCQ

JEE Main 2014 — Electrochemistry Question with Solution

JEE Main 2014 (06 Apr)

Question

Given below are the half - cell reactions :

Mn 2 + + 2 e - → Mn  ;  E ∘ = - 1.18 V 2 Mn 3 + + e - → Mn 2 +  ;  E ∘ = + 1.51 V

The E ∘ for 3 Mn 2 + → Mn + 2 Mn 3 + will be :

Choose an option

Show full solutionCorrect option: A
Correct answer
A - 2.69 V ; the reaction will not occur

Step-by-step explanation

Δ G ∘ = - nF E ∘

Mn + 2 + 2 e - → Mn    ........... (1)

Δ G = - 2 × F × - 1.18 = + 2.36 F

2 Mn + 3 + 2 e - → 2 Mn + 2    ....... (2)

ΔG=-2×F×+1.51=-3.02 F

(1) - (2)

3 Mn + 2 → Mn + 2 Mn + 3

Δ G = + 2.36F +  3.02F = 5.38 F

Δ G = - 2 × F × E

E = 5.38F - 2 × F = - 2.69 V

We know that when Ecell < 0, cell is non-spontaneous.

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About this question

This is a previous-year question from JEE Main 2014, covering the Electrochemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.