At what pH, given half cell MnO_4^-(0.1 M) | Mn^2+(0.001 M) will have electrode potential of 1.282 V? ___________ (Nearest Integer) Given E_MnO_4^ - |Mn^2 + ^o=1.54 V,frac2.303 RT F=0.059 V

MnO_4^-+8 H^++5 e^- Mn^2++4 H_2 O E= E^°-(0.059 / 5) log frac[ Mn^2+][ MnO_4^-][ H^+]^8 \ => 1.282=1.54-(0.059 / 5) log frac10^-310^-1 x [ H^+]^8 \ => (0.258 x 5 / 0.059)=log frac10^-2[ H^+]^8 \ => 21.86=-2+8 pH \ Therefore pH=2.98 \ = 3 \

At what pH given half cell MnO401MMn20001M will have el... | null (Chemistry)

\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}

Explanation

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