Question

At what $pH$, given half cell $MnO _4^{-}(01 M ) \mid Mn ^{2+}(0001 M )$ will have electrode potential of $1282\, V$? (Nearest Integer)
(Given: $E _{ MnO _{4}^-| Mn ^{2+}}^{ o }=154 \,V , \frac{2303 RT }{ F }=0059\, V$)

Answer 1

MnO4−​+8H++5e−⇌Mn2++4H2​O
E=E∘−50.059​log[MnO4−​][H+]8[Mn2+]​
1.282=1.54−50.059​log10−1×[H+]810−3​
0.0590.258×5​=log[H+]810−2​
⇒21.86=−2+8pH
∴pH=2.98
≃3

So, the correct answer is 3.