Question

Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Ag}\left|\mathrm{Ag}^{+} \| \mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}\right| \mathrm{Pt}\) (b) \(\mathrm{Al}\left|\mathrm{Al}^{3+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) (c) \(\mathrm{Pt}\left|\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+} \| \mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+}\right| \mathrm{Pt}\)

Short answer

Question: Calculate the standard cell potential (E°) for each given cell: (a) Ag+|Ag || Sn4+|Sn2+ (b) Al3+|Al || Cu2+|Cu (c) Fe3+|Fe2+ || MnO4-|Mn2+ Answer: (a) The standard cell potential for the given cell Ag+|Ag || Sn4+|Sn2+ is 0.65 V. (b) The standard cell potential for the given cell Al3+|Al || Cu2+|Cu is 2.00 V. (c) The standard cell potential for the given cell Fe3+|Fe2+ || MnO4-|Mn2+ is 0.74 V.

Step-by-step solution

(a) Identifying the reduction half-cell reactions

First, identify the reduction half-cell reactions for the given cell: \(\mathrm{Ag}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}\) (reduction in left half-cell) \(\mathrm{Sn}^{4+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+}\) (reduction in right half-cell)

(a) Finding the standard potentials

Look up the standard reduction potentials for each half-cell reaction: \(\mathrm{Ag}^{+} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}\), \(E^{\circ}_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.80 \ \text{V}\) \(\mathrm{Sn}^{4+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+}\), \(E^{\circ}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}} = 0.15 \ \text{V}\)

(a) Calculating the cell potential

Use the Nernst equation to calculate the cell potential: \(E^{\circ} = E^{\circ}_{\mathrm{cathode}} - E^{\circ}_{\mathrm{anode}} = E^{\circ}_{\mathrm{Ag}^{+}/\mathrm{Ag}} - E^{\circ}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}} = 0.80 - 0.15 = 0.65 \ \text{V}\)

(b) Identifying the reduction half-cell reactions

First, identify the reduction half-cell reactions for the given cell: \(\mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}\) (reduction in left half-cell) \(\mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}\) (reduction in right half-cell)

(b) Finding the standard potentials

Look up the standard reduction potentials for each half-cell reaction: \(\mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}\), \(E^{\circ}_{\mathrm{Al}^{3+}/\mathrm{Al}} = -1.66 \ \text{V}\) \(\mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}\), \(E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34 \ \text{V}\)

(b) Calculating the cell potential

Use the Nernst equation to calculate the cell potential: \(E^{\circ} = E^{\circ}_{\mathrm{cathode}} - E^{\circ}_{\mathrm{anode}} = E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} - E^{\circ}_{\mathrm{Al}^{3+}/\mathrm{Al}} = 0.34 - (-1.66) = 2.00 \ \text{V}\)

(c) Identifying the reduction half-cell reactions

First, identify the reduction half-cell reactions for the given cell: \(\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\) (reduction in left half-cell) \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\) (reduction in right half-cell)

(c) Finding the standard potentials

Look up the standard reduction potentials for each half-cell reaction: \(\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\), \(E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.77 \ \text{V}\) \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\), \(E^{\circ}_{\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}} = 1.51 \ \text{V}\)

(c) Calculating the cell potential

Use the Nernst equation to calculate the cell potential: \(E^{\circ} = E^{\circ}_{\mathrm{cathode}} - E^{\circ}_{\mathrm{anode}} = E^{\circ}_{\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}} - E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 1.51 - 0.77 = 0.74 \ \text{V}\)