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