In the redox reaction between $\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{H}^{+}$ and sulphite ion, what is the number of moles of electrons involved in producing 3.0 moles of the oxidised product?

The balanced redox reaction between dichromate ion and sulphite ion in acidic medium is given by:

$\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 3\mathrm{SO}_{3}^{2-} + 8\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{SO}_{4}^{2-} + 4\mathrm{H}_{2}\mathrm{O}$

The oxidation half-reaction is the oxidation of sulphite ion to sulphate ion:

$\mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 2\mathrm{H}^{+} + 2e^{-}$

In this half-reaction, 2 moles of electrons are involved in the production of 1 mole of the oxidised product, which is $\mathrm{SO}_{4}^{2-}$.

For the production of 3.0 moles of the oxidised product ($\mathrm{SO}_{4}^{2-}$), the number of moles of electrons involved is:

$3.0 \text{ moles of } \mathrm{SO}_{4}^{2-} \times \dfrac{2 \text{ moles of } e^{-}}{1 \text{ mole of } \mathrm{SO}_{4}^{2-}} = 6.0 \text{ moles of } e^{-}$

Answer: 6