Chemistry 9701 · AS & A Level · Electrochemistry

Electrochemistry — practice question

(a)[1]

Define the term standard cell potential, $E^\circ_{\text{cell}}$.

(b)[3]

An electrochemical cell is arranged to determine $E^\circ_{\text{cell}}$ for a cell made from an $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell and a $\text{Cl}_2/\text{Cl}^-$ half-cell. Draw a labelled diagram of this electrochemical cell. Include every necessary substance. It is not necessary to state the conditions used.

(c)[2]

The cell reaction for the electrochemical cell in (b) is given below. $\text{Cl}_2 + 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2\text{Cl}^-$ $E^\circ_{\text{cell}} = +0.59\,\text{V}$ Calculate $\Delta G^\circ$, in $\text{kJ mol}^{-1}$, for this cell reaction.

(d(i))[1]

A second experiment is carried out with the same electrochemical cell. In this experiment, the $\text{Fe}^{2+}$ concentration is $0.15\,\text{mol dm}^{-3}$. All other concentrations stay at their standard values. The Nernst equation is given below. $E = E^\circ + (0.059/z) \log\left(\frac{\text{oxidised species}}{\text{reduced species}}\right)$ Use the Nernst equation to calculate the electrode potential, $E$, for the $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell in this experiment. $[E^\circ : \text{Fe}^{3+}/\text{Fe}^{2+} = +0.77\,\text{V}]$

(d(ii))[1]

Use the result from (d)(i) to calculate $E_{cell}$ for this electrochemical cell.

Worked solution & mark scheme

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