At what pH, given half cell $\mathrm{MnO_{4}^{-}(0.1~M)~|~Mn^{2+}(0.001~M)}$ will have electrode potential of 1.282 V? ___________ (Nearest Integer)
Given $$\mathrm{E_{MnO_4^ - |M{n^{2 + }}}^o}=1.54~\mathrm{V},\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059\mathrm{V}$$
Answer (integer)
3
Solution
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightleftharpoons \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
<br/><br/>$$
\begin{aligned}
& \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{2+}\right]}{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8} \\\\
& \Rightarrow 1.282=1.54-\frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times\left[\mathrm{H}^{+}\right]^8} \\\\
& \Rightarrow \frac{0.258 \times 5}{0.059}=\log \frac{10^{-2}}{\left[\mathrm{H}^{+}\right]^8} \\\\
& \Rightarrow 21.86=-2+8 \mathrm{pH} \\\\
& \therefore \mathrm{pH}=2.98 \\\\
& \simeq 3 \\\\
&
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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