CAIE A-Level Chemistry 24.2.4 Standard Cell-Potential Calculations

Practise combining two standard reduction potentials to calculate E°cell without multiplying potential values.

Exam points

select the more positive reduction potential as the cathode for the spontaneous cell direction
calculate E°cell = E°cathode − E°anode using both values as tabulated reduction potentials
balance electrons in half-equations but never multiply an E° value by its stoichiometric coefficient

[Maximum number: 3]

Titanium is a transition element in Period 4. It is commonly found as $\mathrm{TiO}_{2}$ in minerals.

Acidified $\mathrm{Ti}^{3+}(\mathrm{aq})$ reacts with oxygen dissolved in water as shown.

4 \mathrm{Ti}^{3+}+\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{TiO}^{2+}+4 \mathrm{H}^{+} \quad \Delta G^{\ominus}=-436.1 \mathrm{~kJ} \mathrm{~mol}^{-1}

The standard reduction potential, $E^{\ominus}$ , of $\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}$ is +1.23 V .

Calculate the standard reduction potential, $E^{\ominus}$ , in V , of the $\mathrm{TiO}^{2+}(\mathrm{aq}) / \mathrm{Ti}^{3+}(\mathrm{aq})$ half-cell. Show your working.

E^{\ominus}=

$\begin{aligned} \text { M1: } \Delta G=-n E \ominus_{\text {cell }} F \text { AND } n=4 \text { M2: } \therefore E \ominus_{\text {cell }}=-436100 /-4(96500)=1.13 \mathrm{~V} \text { ecf } \text { M3: } E \ominus_{\text {cell }}=E \ominus\left(O_{2}, 4 H^{+} \mid \mathrm{H}_{2} \mathrm{O}\right)-E \ominus\left(\mathrm{TiO}^{2+} \mid \mathrm{Ti}^{3+}\right)=1.23-E \ominus\left(\mathrm{TiO}^{2+} \mid \mathrm{Ti}^{3+}\right) \therefore E \ominus\left(\mathrm{TiO}^{2+} \mid \mathrm{Ti}^{3+}\right)=(+) \mathbf{0 . 1}(\mathrm{V}) \text { ecf } \end{aligned}$

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24.2.4—A standard cell potential by combining two question 1