Question

The amount of current in Faraday is required for the reduction of $1$ mol of $C{r}_2{O}_7^{2-}$ ions to $C{r}^{3+}$ is:

• A. $1F$
• B. $2F$
• C. $6F$
• D. $4F$

Answer 1

Answer: (C)

$6F$

Oxidation state of $Cr$ changes from $+6\left({C{r}_2{O}_7}^{2-}\right)$ to $+3\left(C{r}^{3+}\right)$.

Therefore, n-factor of $C{r}_2{O}_7^{2-}$ will be $n_f=2(6-3)=6.$

6 moles of electrons will be required for the reduction of 1 mole of $C{r}_2{O}_7^{2-}$.

i.e. $6F$ charge will be required.

Hence, option C is correct.