Question

$x & y$ are the number of moles of electrons involved respectively during oxidation of $\text{I}^-$ to $\text{I}_2$ & $\text{S}^{2-}$ to $\text{S}$ by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of $x+y$ is ?

• A. 12
• B. 6
• C. 18
• D. 9

Hint:

When asked for the number of electrons involved in an oxidation step catalyzed by a specific amount of oxidant, calculate the total electrons required by the oxidant ($\text{Cr}_2\text{O}_7^{2-}$ requires $6\text{e}^-$) and scale the oxidation half-reaction accordingly.

Answer 1

The Correct Option is A

The reduction of $1 \text{ mole of } \text{Cr}_2\text{O}_7^{2-}$ requires $6 \text{ moles of electrons}$.
$x$: Moles of electrons involved during oxidation of $\text{I}^- \rightarrow \text{I}_2$.
$3 \times (2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-) \implies 6\text{e}^-$. $x=6$.
$y$: Moles of electrons involved during oxidation of $\text{S}^{2-} \rightarrow \text{S}$.
$3 \times (\text{S}^{2-} \rightarrow \text{S} + 2\text{e}^-) \implies 6\text{e}^-$. $y=6$.
Sum of $x+y = 6 + 6 = 12$.