At what pH , given half cell MnO _ 4 ^ - ( 0 . 1M ) ∣ Mn ^ 2 + ( 0...

${MnO}_{4}^{-} + 8{H}^{+} + 5{e}^{-} ⇌ {Mn}^{2 +} + 4{H}_{2}O$

$E = E ° - \dfrac{0 . 059}{5}log\dfrac{\left[{Mn}^{2 +}\right]}{\left[{MnO}_{4}^{-}\right]{\left[{H}^{+}\right]}^{8}}$

$1 . 282 = 1 . 54 - \dfrac{0 . 059}{5}log\dfrac{{10}^{- 3}}{{10}^{- 1} \times {\left[{H}^{+}\right]}^{8}}$

$\dfrac{0 . 258 \times 5}{0 . 059} = log\dfrac{{10}^{- 2}}{{\left[{H}^{+}\right]}^{8}}$

$⇒ 21 . 86 = - 2 + 8pH$

$pH = 2 . 98$

$≃ 3$