Question

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

• A. 8
• B. 2
• C. 3
• D. 6

Hint:

When dealing with redox reactions, make sure to balance the oxidation states of all elements involved to determine the number of electrons transferred.

Answer 1

The Correct Option is D

In a redox reaction, the number of electrons involved in the reaction can be determined by looking at the changes in oxidation states of the reacting species. 
In the reaction between \( \text{Cr}_2\text{O}_7^{2-} \) and the sulphite ion \( \text{SO}_3^{2-} \), chromium (Cr) undergoes a change in oxidation state, and the number of electrons is involved in its reduction. 
The balanced half-reaction for chromium in acidic medium is: \[ \text{Cr}_2\text{O}_7^{2-} + 14 \, \text{H}^+ + 6 \, \text{e}^- \rightarrow 2 \, \text{Cr}^{3+} + 7 \, \text{H}_2\text{O} \] The reduction half-reaction shows that for each mole of \( \text{Cr}_2\text{O}_7^{2-} \), 6 moles of electrons are involved. Since the problem asks for the number of electrons involved in producing 3.0 moles of the oxidised product, we multiply the number of electrons involved per mole of \( \text{Cr}_2\text{O}_7^{2-} \) by 3: \[ 6 \, \text{e}^- \times 3 = 6 \, \text{e}^- \] Thus, the number of moles of electrons involved is \( 6 \, \text{mol} \).