Verified Solution • 2025–26 Session moderate

QJEEM23Chdef6: JEE Main 2023 Chemistry - Class 12 JEEM23Chdef6 Solution: At what pH, given half cell will have electrode potential of…

At what pH, given half cell MnO4−(0.1 M) ∣ Mn2+(0.001 M)\mathrm{MnO_{4}^{-}(0.1~M)~|~Mn^{2+}(0.001~M)} will have electrode potential of 1.282 V? ___________ (Nearest Integer)

Given EMnO4−∣Mn2+o=1.54 V,2.303RTF=0.059V\mathrm{E_{MnO_4^ - |M{n^{2 + }}}^o}=1.54~\mathrm{V},\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059\mathrm{V}

Step-by-Step Verified Solution

MnO4−+8H++5e−⇌Mn2++4H2O\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightleftharpoons \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}

amp;E=E∘−0.0595log⁡[Mn2+][MnO4−][H+]8amp;⇒1.282=1.54−0.0595log⁡10−310−1×[H+]8amp;⇒0.258×50.059=log⁡10−2[H+]8amp;⇒21.86=−2+8pHamp;∴pH=2.98amp;≃3amp;\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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