Question

Charge required to reduce one mole of $C{r}_2{O}_7^{2-}$ to $C{r}^{3+}$ in an acidic medium is:

• A. 3 Faraday
• B. 6 Faraday
• C. 0.3 Faraday
• D. 0.6 Faraday

Answer 1

The correct option is B 6 Faraday

For the reaction of $C{r}_2{O}_7^{2-}$ in acidic medium,
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$

Here, for the reduction of one mole of $C{r}_2{O}_7^{2-}$ six moles of $e^{-}$ are required.

$1mol\text{of}{e}^{-}=1\text{Faraday of charge}$
$\therefore$
Charged required for reduction of one mole of $C{r}_2{O}_7^{2-}$ is,
$\text{Charge required}=6\times 1Faraday\Rightarrow 6F$