Question

By how much would the oxidizing power of the $Mn{O}_4^{-}/M{n}^{2+}$ couple change if the $H^{+}$ ions concentration is decreased to a hundredth of its initial value?

• A. Increases by 189 mV
• B. Decreases by 189 mV
• C. Increases by 19 mV
• D. Decreases by 19 mV

Answer 1

The correct option is B

Decreases by 189 mV

$Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$

According to Nernst equation,

$E_{red}=E_{red}^{\circ }-\frac{0.059}{5}log\left\{\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right][H^{+}{]}^8}\right\}$

Let the initial $\left[H^{+}\right]=X$

$E_{red\_init}=E_{red}^{\circ }-\frac{0.059}{5}log\left\{\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]X^8}\right\}$

$E_{red\_final}=E_{red}^{\circ }-\frac{0.059}{5}log\left\{\frac{\left[M{n}^{2+}\right]\times {10}^{16}}{\left[Mn{O}_4^{-}\right]X^8}\right\}$

$\therefore E_{red\_final}-E_{red\_init}=-\frac{0.059}{5}log\left\{{10}^{16}\right\}=-0.189V$

The tendency of the half cell to get reduced is its oxidising power. Hence the oxidising power decreases by 189mV