Question

The magnitude of the change in oxidising power of the $Mn{O}_4^{-}/M{n}^{2+}$ couple is $x\times {10}^{-4}V$, if the $H^{+}$ concentration is decreased from $1M$ to ${10}^{-4}M$ at $25°C$. (Assume concentration of $Mn{O}_4^{-}$ and $M{n}^{2+}$ to be same on change in $H^{+}$ concentration). The value of $x$ is . (Round off to the nearest integer).

[Given: $2.30RT/F=0.059$]

Answer 1

$5e^{-}+8H^{+}+\underset{1M}{Mn{O}_4^{-}}\to \underset{1M}{M{n}^{2+}}+4H_{2}O$

$E_1=E^o-\frac{0.059}{5}{\log }_{10}[\frac{1}{{\left[H^{+}\right]}^8}\times \frac{\left[M{n}^{+2}\right]}{[Mn{O}_4^{-}}]$

$=E^o-\frac{0.059}{5}{\log }_{10}\left[\frac{1}{{\left(1\right)}^8}\right]=E^o$

$E_2=E^o-\frac{0.059}{5}{\log }_{10}[\frac{1}{{\left({10}^{-4}\right)}^8}\times \frac{\left[M{n}^{+2}\right]}{\left[Mn{O}_4^{-}\right]}]$

$=E^o-\frac{0.059}{5}{\log }_{10}\left[{10}^{32}\right]$

$=E^o-\frac{0.059}{5}\times 32$
$E_2-E_1=\frac{0.059}{5}\times (-32)|(E_2-E_1)|=\frac{0.059}{5}\times \left(32\right)=x\times {10}^{-4}$
$\frac{32\times 590}{5}\times {10}^{-4}=x\times {10}^{-4}$
$=3776\times {10}^{-4}$
$x=3776$