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The charge required for reducing 1 mole ...

The charge required for reducing 1 mole of `MnO_4^(-)` to `Mn^(2+)` is

A

`1.93xx10^5` C

B

`2.895xx10^5` C

C

`4.28xx10^5` C

D

`4.825xx10^5` C

Text Solution

Verified by Experts

The correct Answer is:
D
`underset"1mol"(MnO_4^-) + underset"5 moles "(5e^(-)) to underset"1 mole"(Mn^(2+))`
5 moles of electrons are needed for reduction of 1 mole of `MnO_4^-` to `Mn^(2+)`
5 moles of electrons = 5 Faradays
Quantity of charge required = 5 x 96500 = 4.825 x `10^5` Coulombs
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