We need to find the number of electrons (X) involved when iodide ion ($\text{I}^-$) is oxidized to iodine ($\text{I}_2$) by acidified $\text{K}_2\text{Cr}_2\text{O}_7$.
The oxidation half-reaction is: $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$
However, the question asks for electrons involved *during the oxidation... by acidified $\text{K}_2\text{Cr}_2\text{O}_7$*. This implies balancing the electrons transferred in the oxidation half-reaction against the reduction of the oxidizing agent, $\text{K}_2\text{Cr}_2\text{O}_7$.
The reduction half-reaction for dichromate ($\text{Cr}_2\text{O}_7^{2-}$) in acidic medium is: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ This reaction involves the gain of 6 electrons.
To balance the 6 electrons gained by dichromate, the oxidation of iodide must supply 6 electrons. The simple half-reaction $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$ needs to be multiplied by 3 to provide 6 electrons: $3 \times (2\text{I}^- \rightarrow \text{I}_2 + 2e^-)$ $6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-$ Therefore, the number of electrons involved in the oxidation of iodide in this context is X = 6.
We need to find the number of electrons (Y) involved when sulfide ion ($\text{S}^{2-}$) is oxidized to elemental sulfur (S) by acidified $\text{K}_2\text{Cr}_2\text{O}_7$.
The oxidation half-reaction is: $\text{S}^{2-} \rightarrow \text{S} + 2e^-$
Similar to the iodide case, we need to consider the electron transfer relative to the reduction of dichromate (which involves 6 electrons).
To balance the 6 electrons gained by dichromate, the oxidation of sulfide must supply 6 electrons. The half-reaction $\text{S}^{2-} \rightarrow \text{S} + 2e^-$ needs to be multiplied by 3 to provide 6 electrons: $3 \times (\text{S}^{2-} \rightarrow \text{S} + 2e^-)$ $3\text{S}^{2-} \rightarrow 3\text{S} + 6e^-$ Therefore, the number of electrons involved in the oxidation of sulfide in this context is Y = 6.
We found that X = 6 and Y = 6.
The total number of electrons is the sum: $ \text{X} + \text{Y} = 6 + 6 = 12 $
| $\text{c } [\text{mol L}^{-1}]$ | $0.04$ | $0.09$ | $0.16$ | $0.25$ |
| $\Lambda_{\text{m}} \ [\text{S cm}^{2}\text{ mol}^{-1}]$ | $96.1$ | $95.7$ | $95.3$ | $94.9$ |
For the thermal decomposition of reactant $\text{AB(g)}$, the following plot is constructed. 
The half life of the reaction is '$x$' min.
$x = $ ___________ min. (Nearest integer)