# The magnitude of the change in oxidising power of the

Question:

The magnitude of the change in oxidising power of the $\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}$ couple is $\mathrm{x} \times 10^{-4} \mathrm{~V}$, if the $\mathrm{H}^{+}$concentration is decreased from $1 \mathrm{M}$ to $10^{-4} \mathrm{M}$ at $25^{\circ} \mathrm{C}$. (Assume concentration of $\mathrm{MnO}_{4}^{-}$and $\mathrm{Mn}^{2+}$ to be same on change in $\mathrm{H}^{+}$ concentration). The value of $x$ is (Rounded off to the nearest integer) $\left[\right.$ Given $\left.: \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059\right]$

Solution:

Eqn is-

$\mathrm{MnO}_{4}^{-}+\mathrm{H}^{\oplus}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}$

Nernst equation:

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {Cell }}^{0}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{+2}\right]}{\left[\mathrm{MnO}_{4}^{-}\right]}\left[\frac{1}{\mathrm{H}^{+}}\right]^{8}$

(I) Given $\left[\mathrm{H}^{\oplus}\right]=1 \mathrm{M}$

$\mathrm{E}_{1}=\mathrm{E}^{0}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{+2}\right]}{\left[\mathrm{MnO}_{4}^{-}\right]}$

(II) Now : $\left[\mathrm{H}^{\oplus}\right]=10^{-4} \mathrm{M}$

$\mathrm{E}_{2}=\mathrm{E}^{0}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{+2}\right]}{\left[\mathrm{MnO}_{4}^{-}\right]} \times \frac{1}{\left(10^{-4}\right)^{8}}$

$=\mathrm{E}^{0}-\frac{0.059}{5} \log \frac{\mathrm{Mn}^{+2}}{\left[\mathrm{MnO}_{4}^{-}\right]}+\frac{0.059}{5} \log 10^{-32}$

therefore : $\left|E_{1}-E_{2}\right|=\frac{0.059}{5} \times 32$

$=0.3776 \mathrm{~V}=3776 \times 10^{-4}$

$x=3776$