The number of electrons lost during the oxidation of `Cr^(3+) " to " Cr_(2)O_(7)^(2-)` are

Answer

3

Explanation

`7H_2O+2Cr^(3+)rarrCr_(2)O_(7)^(2-)+14H^(+)+6e^(-)`
1 mol of `Cr^(3+)` will loose 3 mol electrons while oxidised to `Cr_(2)O^(2-)`