Redox and electrochemistry — Question 138

Answer
Cell A: $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{N}\mathrm{a}$ and $2\mathrm{B}{\mathrm{r}}^{-}\to \mathrm{B}{\mathrm{r}}_2+2{\mathrm{e}}^{-}$. Cell B forms H₂ at the cathode and O₂ at the anode. Water, rather than Na⁺, is reduced; the H₂:O₂ mole ratio is 2:1.

Working
Molten sodium bromide has Na⁺ and Br⁻ available but no water. In cell B, water reduction is $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2\mathrm{O}{\mathrm{H}}^{-}$ and water oxidation is $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$. Matching four electrons gives two H₂ per O₂. The supplied aqueous model selects water reduction instead of sodium-ion reduction.

Marking criteria

• Writes the balanced sodium-ion reduction half-equation. [1 mark]
• Writes the balanced bromide oxidation half-equation. [1 mark]
• Identifies hydrogen at the cathode and oxygen at the anode in cell B. [1 mark]
• Explains preferential water reduction rather than sodium-ion reduction in aqueous solution. [1 mark]
• Determines a 2:1 cathode/anode gas mole ratio. [1 mark]