CAIE A-Level Chemistry 24.2.5 Cell Potentials and Electron Flow

Practise using E°cell to assign electrode polarity, electron-flow direction and reaction feasibility.

Exam points

identify the oxidation half-cell as the negative electrode and the reduction half-cell as positive
state that electrons flow externally from the negative oxidation electrode to the positive reduction electrode
predict the written reaction is feasible under standard conditions when its calculated E°cell is positive

[Maximum number: 1]

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 .

Acidified $\left[\mathrm{Ti}\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)_{2}\right]^{3-}(\mathrm{aq})$ does not react with oxygen dissolved in water, unlike acidified $\mathrm{Ti}^{3+}(\mathrm{aq})$ .

Suggest what this means for the value of the standard reduction potential, $E^{\ominus}$ , of the following half-cell.

\left[\mathrm{Ti}\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)_{2}\right]^{2-}(\mathrm{aq})+\mathrm{e}^{-} \rightleftharpoons\left[\mathrm{Ti}\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)_{2}\right]^{3-}(\mathrm{aq})

Explain your answer.

the $E \ominus$ of the half-cell must be greater than $+1.23 \mathrm{~V} / E^{\ominus}$ of the $\mathrm{O}_{2} \mid \mathrm{H}^{+}$ half-cell
as $E_{\text {cell }}<0$ and the reaction does not occur

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24.2.5—Standard cell potentials question 1