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Electrochemistry question

2013 · Q88
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Electrochemistry question

2013 · Q88

NEETChemistryElectrochemistryMCQ+4 / −1
Consider the half-cell reduction reaction
Mn2+ + 2e−-− →\to→ Mn, Eo = −-−1.18 V
Mn2+ →\to→ Mn3+ + e−-−, Eo = −-− 1.51 V
The EEEo for the reaction 3 Mn2+ →\to→ Mno + 2Mn3+, and possibility of the forward reaction are respectively
  1. A
    −-− 4.18 V and yes
  2. B
    + 0.33 V and yes
  3. C
    + 2.69 V and no
  4. D
    −-− 2.69 V and no
View written solutionFree

Correct answer: D

Mn2+ + 2e−-− →\to→ Mn, Eo = −-−1.18 V

Mn2+ →\to→ Mn3+ + e−-−, Eo = −-− 1.51 V

3Mn2+ →\to→ Mno + 2Mn3+, Δ\Delta ΔEo = - 1.81 - 1.51 = −-− 2.69 V

Since Δ\Delta ΔE° is negative,

∴\therefore∴ Δ\Delta ΔG = –nFE°, Δ\Delta ΔG will have positive value so, forward reaction is not possible.

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