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STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electrochemistry4 Marks
MCQ
Consider the half-cell reduction reaction

$Mn^{2+} + 2e^- \rightarrow Mn,\, $$E^o = - 1.18\, V$

$Mn^{2+} \rightarrow  Mn^{3+} + e^-,$ $ E^o = - 1.51 \,V$

The $E^o$ for the reaction $3Mn^{2+} \rightarrow  Mn^o + 2Mn^{3+},$

and possibility of the forward reaction are respectively

  • A
    $- 4.18\, V$ and yes
  • B
    $+ 0.33\, V$ and yes
  • C
    $+ 2.69\, V$ and no
  • ✓
    $- 2.69\, V$ and no
✓

Answer

Correct option: D.
$- 2.69\, V$ and no
d
$\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}, \mathrm{E}^{\circ}=-1.18 \mathrm{V}$ .... $(i)$

$2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\right), E^{\circ}=+1.51 \mathrm{V}$.... $(ii)$

Subtracting Eq. $(ii)$ from Eq. $(i)$, we get

${3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}}$

${E^{\circ}=-1.18-(+1.51)=-2.69 \mathrm{V}}$

since, the value of $E^{\circ}$ is -ve, therefore the reaction is non-spontaneous

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