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We consider the following redox system: $MnO_4^{-}$ // $Mn^{2+}$ $E^o$= $1.52 \;V$ What is his → potential under the standard conditions and at pH=1 for $[MnO_4^{-}]$ $=$ $0.1 \frac{mol}{L}$ and $[Mn^{2+}]$ $=$ $0.1 \frac{mol}{L}$?
$MnO_4^{-}$ + 5 $e^- $ +$8H^+$ $\leftrightarrows$ $Mn^{2+}+4H_2O$
$E$ $ =$ $ E^o+\frac{0.059}{n}log\frac{[MnO_4^{-}]\cdot[H^+]^8}{[Mn^{2+}]}$
$n\; =\; 5$
$E$ $ =$ $ 1.52+\frac{0.059}{5}log(\frac{0.1\cdot 10^{-8}}{0.1})$ $ =$ $1.43\;V$