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Q. $MnO^{-}_{4} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_{2}O$
If $H^{+}$ concentration is decreased from 1 M to $10^{-4}$ M at $25°C$,

Electrochemistry

Solution:

$\Delta E =\frac{0.059}{5}log \left[10^{-4}\right]^{8}=-0.378\, V$