Question

Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Pb}\left|\mathrm{PbSO}_{4} \| \mathrm{Pb}^{2+}\right| \mathrm{Pb}\) (b) \(\mathrm{Pt}\left|\mathrm{Cl}_{2}\right| \mathrm{ClO}_{3}^{-} \| \mathrm{O}_{2}\left|\mathrm{H}_{2} \mathrm{O}\right| \mathrm{Pt}\) (c) \(\mathrm{Pt}\left|\mathrm{OH}^{-}\right| \mathrm{O}_{2} \| \mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-} \mid \mathrm{Pt} \quad\) (basic medium)

Short answer

Question: Calculate the standard cell potential, E°, for the given cell: $\mathrm{Pb}\left|\mathrm{PbSO}_{4} \| \mathrm{Pb}^{2+}\right| \mathrm{Pb}$. Answer: Using the provided steps, we have found \(E^{\circ}_1=0.356\,\mathrm{V}\) and \(E^{\circ}_2=-0.126\,\mathrm{V}\). To calculate the overall E° for the cell, we add the two half-reaction potentials: \(E^{\circ}=0.356\,\mathrm{V} -0.126\,\mathrm{V}=0.230\,\mathrm{V}\). The standard cell potential for the given cell is \(0.230\,\mathrm{V}\).

Step-by-step solution

Identify the half-reactions

First, we should identify the half-reactions involved in the given cell: 1. \(\mathrm{PbSO}_{4 \left(solid \right)} + 2e^{-} \rightarrow \mathrm{Pb}^{2+} + \mathrm{SO}^{2-}_{4}\) 2. \(\mathrm{Pb}^{2+} + 2e^{-} \rightarrow \mathrm{Pb}\)

Look for E° values for each half-reaction in a standard reduction potential table

To find the E° values, we must look up the standard reduction potentials for each half-reaction in a table: 1. \(\mathrm{PbSO}_{4 \left(solid \right)} + 2e^{-} \rightarrow \mathrm{Pb}^{2+} + \mathrm{SO}^{2-}_{4}\), \(E^{\circ}_1=0.356\,\mathrm{V}\) (this is the reversal of the given half-reaction, so remember to change the sign) 2. \(\mathrm{Pb}^{2+} + 2e^{-} \rightarrow \mathrm{Pb}\), \(E^{\circ}_2=-0.126\,\mathrm{V}\)

Calculate overall E° for the cell

Add the two half-reaction potentials to get the overall cell potential: \(E^{\circ}=E^{\circ}_1+E^{\circ}_2\)