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Electrochemistry question

2015 · Shift 1 · Q3
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Electrochemistry question

2015 · Shift 1 · Q3

JEE AdvancedChemistryElectrochemistryNumerical+4 / −1
All the energy released from the reaction X→Y,ΔtGoX \to Y, \Delta _tG^oX→Y,Δt​Go= -193 kJ mol-1 is used for oxidizing M+{}^++ as M+{}^++→\to→ M3+{}^{3+}3+ + 2e-, Eo = -0.25 V Under standard conditions, the number of moles of M+{}^++ oxidized when one mole of X is converted to Y is [F = 96500 C mol–1]
Numerical answer
View written solutionFree

Correct answer: 4

  1. Use the relation between Gibbs free energy and electrical work

Under standard conditions,

ΔG∘=−nFE∘\Delta G^\circ = -n F E^\circΔG∘=−nFE∘

For a process, the maximum useful electrical work available is:

−ΔG∘-\Delta G^\circ−ΔG∘

Given for

X→Y,ΔG∘=−193 kJ mol−1X \to Y,\quad \Delta G^\circ = -193\,\text{kJ mol}^{-1}X→Y,ΔG∘=−193kJ mol−1

So energy released by conversion of 1 mole of XXX is

193 kJ=193000 J193\,\text{kJ} = 193000\,\text{J}193kJ=193000J
  1. Find Gibbs free energy required for oxidation of M+M^+M+ to M3+M^{3+}M3+

Given oxidation half-reaction:

M+→M3++2e−M^+ \to M^{3+} + 2e^-M+→M3++2e−

with

E∘=−0.25 VE^\circ = -0.25\,\text{V}E∘=−0.25V

Number of electrons transferred:

n=2n = 2n=2

Hence,

ΔG∘=−nFE∘=−(2)(96500)(−0.25)\Delta G^\circ = -nFE^\circ = -(2)(96500)(-0.25)ΔG∘=−nFE∘=−(2)(96500)(−0.25) ΔG∘=48250 J mol−1=48.25 kJ mol−1\Delta G^\circ = 48250\,\text{J mol}^{-1} = 48.25\,\text{kJ mol}^{-1}ΔG∘=48250J mol−1=48.25kJ mol−1

This is positive, so oxidation requires energy input of 48.25 kJ48.25\,\text{kJ}48.25kJ per mole of M+M^+M+ oxidized.

  1. Calculate moles of M+M^+M+ oxidized using total released energy

If all 193 kJ193\,\text{kJ}193kJ released is used, then number of moles oxidized is

19348.25=4\frac{193}{48.25} = 448.25193​=4
  1. Final answer

The number of moles of M+M^+M+ oxidized is

4\boxed{4}4​
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