How much charge is required for the following reductions: 1 mol of `MnO_4^(-)` to Mn2+.
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Solution
`MnO_4^(-) -> Mn^(2+)`
ie `Mn^(7+) + 5e^(-) -> Mn^(2+)`
∴ Required charge = 5 F
= 5 × 96487 C
= 482435 C
Concept: Electrochemical Cells
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