The charge required for the reduction of 1 mol of `Cr_(2)O_(7)^(2-)` to `Cr^(3+)` is :

Options

A. 96500 C
B. `1.93xx10^(5)C`
C. `5.79xx10^(5)C`
D. `2.895xx10^(5)C`

Answer

C

Explanation

`Cr_(2)O_(7)^(2-) +6e^(-) rarr 2Cr^(3+)`
Reduction of 1 mol of `Cr_(2)O_(7)^(2-)` to `Cr^(3+)` requires 6 mol of electrons
`:.` Charge required `=6xx96500C`
`=5.79xx10^(5)C`