How many electrons are transferred per dichromate ion when $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ is reduced to $\mathrm{Cr}^{3+}$ in acidic medium?

1. The half-reaction in acidic medium is: $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\,\mathrm{H}^+ + n\,\mathrm{e}^- \rightarrow 2\,\mathrm{Cr}^{3+} + 7\,\mathrm{H_2O}$. 2. Track Cr oxidation states. In $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, Cr is $+6$ (from $2x + 7(-2) = -2$, so $x = +6$). In $\mathrm{Cr}^{3+}$, Cr is $+3$. 3. Each Cr gains $6 - 3 = 3$ electrons. There are 2 Cr atoms, so total electrons gained = $2 \times 3 = 6$. 4. Verify by charge balance: LHS charge = $(-2) + 14(+1) + 6(-1) = +6$; RHS charge = $2(+3) + 7(0) = +6$. ✓ 5. Also verify atom balance: Cr 2=2, O 7=7, H 14=14. ✓ The balanced half-reaction is the textbook acidic-medium half-reaction used in volumetric titrations. _Source: syllabus Chemistry Part 2, Ch 7 "Redox Reactions", §7.4 (Balancing of Redox Reactions — half-reaction method, dichromate example), p. 12–13._