Consider the following reaction
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^ -$ is ___________. (Integer answer)
Answer (integer)
25
Solution
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$
<br><br>1 mole of MnO<sub>4</sub><sup>-</sup> required 5 moles of
electrons or 5 F electricity.
<br><br>$\therefore$ 5 moles of MnO<sub>4</sub><sup>-</sup> required 25 F electricity.
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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