Redox equilibria

WCH15Topic 5.13 questions

Standard electrode potentials, electrochemical cells, redox titrations and fuel cells.

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Question 13 marks
A hydrogen–oxygen fuel cell uses an alkaline electrolyte.

Write the half-equation for the reaction at each electrode and the overall equation for the cell.
Mark scheme
- Negative electrode: $\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2\text{e}^-$ (1)
- Positive electrode: $\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightarrow 4\text{OH}^-$ (1)
- Overall: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ (1)
Question 25 marks
A $0.250\,\text{g}$ sample of crushed iron tablets was dissolved in dilute sulfuric acid, so that all the iron was present as $\text{Fe}^{2+}$ . The solution was titrated with $0.0100\,\text{mol dm}^{-3}$ potassium manganate(VII); was needed.

−

→

4OH−

−3

18.40\,\text{cm}^3

$\text{MnO}_4^{\;-} + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

Calculate the percentage by mass of iron in the tablets. [ $A_r$ Fe $= 55.8$ ]

Mark scheme

$n(\text{MnO}_4^{\;-}) = 0.01840 \times 0.0100 = 1.84 \times 10^{-4}\,\text{mol}$ (1)
$n(\text{Fe}^{2+}) = 5 \times 1.84 \times 10^{-4} = 9.20 \times 10^{-4}\,\text{mol}$ (1)
Mass of iron $= 9.20 \times 10^{-4} \times 55.8 = 0.0513\,\text{g}$ (1)
Percentage $= 0.0513 \div 0.250 \times 100$ (1)
$= 20.5\%$ (1)

9.20×

10−4mol