Question

If the $H ^{+}$concentration is decreased from $1\, M$ to $10^{-4} M$ at $25^{\circ} C$ for the couple $MnO _{4}^{-} / Mn ^{2+}$, then the oxidising power of the $MnO _{4}^{-} / Mn ^{2+}$ couple decreases by:

• A. -0.18 V
• B. 0.18 V
• C. 0.38 V
• D. -0.38 V

Answer 1

Answer: (C)

Initial
$E_{1}=E^{\circ}-\frac{0.0591}{5} \log \frac{\left[ Mn ^{2+}\right]}{(1)^{8}}$
final
$E _{2}= E ^{\circ}-\frac{0.0591}{5} \log \frac{\left[ Mn ^{2+}\right]}{\left(10^{-4}\right)^{8}}$
$E _{2}- E _{1}=\frac{0.0591}{5} \log \frac{\left[ Mn ^{2+}\right] \times\left(10^{-4}\right)^{8}}{\left[ Mn ^{2+}\right]}$
$\Delta E =\frac{0.0591}{5} \log 10^{-32}$
$=0.378 \sim(0.38\, V )$
decreases by $0.378\, V$
Reaction is $MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O$