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A solution containing $4.5 \mathrm{mM}$ of $\mathrm{MnO}_4^{-}$and $15 \mathrm{mM}$ of $\mathrm{Mn}^{2+}$ shows $\mathrm{pH}$ of 2 . The potential of half-cell reaction is ...... .
(Given, $\log 15=1.76$ and $\log 45=1.653$ and standard potential of $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$ ) is $1.51 \mathrm{~V}$ )
Options:
(Given, $\log 15=1.76$ and $\log 45=1.653$ and standard potential of $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$ ) is $1.51 \mathrm{~V}$ )
Solution:
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Verified Answer
The correct answer is:
$1.31 \mathrm{~V}$
$$
\begin{aligned}
& \text { (b) } \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\
& E=E_{\mathrm{Mn}^{\circ}}{ }^{7+} \mathrm{Mn}^{2+} /+\frac{0.059}{5} \log \frac{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8}{\left[\mathrm{Mn}^{2+}\right]} \\
& E=1.51+\frac{0.059}{5} \log \frac{\left[\frac{4.5}{1000}\right]\left[10^{-2}\right]^8}{\left[\frac{15}{1000}\right]} \\
& \Rightarrow \quad E=1.31 \mathrm{~V}
\end{aligned}
$$
Hence, the correct option is (2).
\begin{aligned}
& \text { (b) } \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\
& E=E_{\mathrm{Mn}^{\circ}}{ }^{7+} \mathrm{Mn}^{2+} /+\frac{0.059}{5} \log \frac{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8}{\left[\mathrm{Mn}^{2+}\right]} \\
& E=1.51+\frac{0.059}{5} \log \frac{\left[\frac{4.5}{1000}\right]\left[10^{-2}\right]^8}{\left[\frac{15}{1000}\right]} \\
& \Rightarrow \quad E=1.31 \mathrm{~V}
\end{aligned}
$$
Hence, the correct option is (2).
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