By how much would the oxidising power of the MnO^4- /Mn²⁺ couple change if the H⁺ ions concentration is decreased 100 times?

(a) increases by 189 mV

(b) decreases by 189 mV

(c) will increase by 19 mV

(d) will decrease by 19 mV

Answers (1)

$\begin{array}{l} \mathrm{MnO}_{4}^{-}+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \\ \text { According to Nernst equation, } \\ \mathrm{E}_{\mathrm{red}}=\mathrm{E}_{\mathrm{red}}^{\circ}-\frac{0.059}{5} \log \left[\frac{\left[\mathrm{Mn}^{2+}\right]}{\left[\mathrm{MnO}_{4}^{-}\right]\left[\mathrm{H}^{+}\right]^{8}}\right] \ \ \ \ \ \ \ \mid \ \ \ \ \ \ \ \ \ \text { Let }\left[\mathrm{H}^{+}\right]_{\text {initial }}=\mathrm{X} \end{array}$

$\begin{aligned} &\mathrm{E}_{\text {red }(\text { initial })}=\mathrm{E}_{\text {red }}^{\mathrm{o}} \frac{-0.0591}{5} \log \left[\frac{\left[\mathrm{Mn}^{2+}\right]}{\left[\mathrm{MnO}_{4}^{-}[\mathrm{X}]^{8}\right.}\right] \mid \quad\left[\mathrm{H}^{+}\right] \text {linal }=\frac{\mathrm{X}}{100}=\frac{\mathrm{X}}{10^{2}}\\ &\mathrm{E}_{\mathrm{red}(\text { final })}=\mathrm{E}_{\mathrm{red}}^{\circ} \frac{-0.0591}{5} \log \frac{\left[\mathrm{Mn}^{2+}\right] \times 10^{16}}{\left[\mathrm{MnO}_{4}^{-}\right] \times[\mathrm{X}]^{8}}\\ &\mathrm{E}_{\text {red }(\text { final })}-\mathrm{E}_{\text {red }(\text { initial })}=\frac{-0.0591}{5} \log 10^{16}=-0.1891 \mathrm{V} \end{aligned}$

$\\\text{This E red decreases by 0.189 V. The tendency of the half cell to get }\\ \text{reduced is its oxidising power. Hence the oxidising power decreases by } 0.189 \mathrm{V}$

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By how much would the oxidising power of the MnO4- /Mn2+ couple change if the H+ ions concentration is decreased 100 times? (a) increases by 189 mV (b) decreases by 189 mV (c) will increase by 19 mV (d) will decrease by 19 mV

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By how much would the oxidising power of the MnO^4- /Mn²⁺ couple change if the H⁺ ions concentration is decreased 100 times?

(a) increases by 189 mV

(b) decreases by 189 mV

(c) will increase by 19 mV

(d) will decrease by 19 mV