Question

If the $H^{+}$concentration is decreased from $1M$ to $10^{-4}M$ at $25^{\circ }C$ for the couple $Mn{O}_4^{-}/M{n}^{2+}$, then the oxidising power of the $Mn{O}_4^{-}/M{n}^{2+}$ couple decreases by:

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

Answer 1

Answer: (C)

Initial
$E_1=E^{\circ }-\frac{0.0591}{5}\log \frac{\left[M{n}^{2+}\right]}{(1{)}^8}$
final
$E_2=E^{\circ }-\frac{0.0591}{5}\log \frac{\left[M{n}^{2+}\right]}{{\left(10^{-4}\right)}^8}$
$E_2-E_1=\frac{0.0591}{5}\log \frac{\left[M{n}^{2+}\right]\times {\left(10^{-4}\right)}^8}{\left[M{n}^{2+}\right]}$
$\Delta E=\frac{0.0591}{5}\log 10^{-32}$
$=0.378\sim (0.38V)$
decreases by $0.378V$
Reaction is $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$