JEE MAIN - Chemistry (2023 - 1st February Morning Shift - No. 14)
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
3
Explanation
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightleftharpoons \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
$$ \begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{2+}\right]}{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow 1.282=1.54-\frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow \frac{0.258 \times 5}{0.059}=\log \frac{10^{-2}}{\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow 21.86=-2+8 \mathrm{pH} \\\\ & \therefore \mathrm{pH}=2.98 \\\\ & \simeq 3 \\\\ & \end{aligned} $$
$$ \begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{5} \log \frac{\left[\mathrm{Mn}^{2+}\right]}{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow 1.282=1.54-\frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow \frac{0.258 \times 5}{0.059}=\log \frac{10^{-2}}{\left[\mathrm{H}^{+}\right]^8} \\\\ & \Rightarrow 21.86=-2+8 \mathrm{pH} \\\\ & \therefore \mathrm{pH}=2.98 \\\\ & \simeq 3 \\\\ & \end{aligned} $$
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