Write an equation to represent the oxidation of \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) by \(\mathrm{PbO}_{2}(\mathrm{s})\) in an acidic solution. Will this reaction occur spontaneously in the forward direction if all other reactants and products are in their standard states and (a) \(\left[\mathrm{H}^{+}\right]=6.0 \mathrm{M} ;\) (b) \(\left[\mathrm{H}^{+}\right]=1.2 \mathrm{M}\) (c) \(\mathrm{pH}=4.25 ?\) Explain.

First, identify the reducing agent (RA, which gets oxidized) and the oxidizing agent (OA, which gets reduced). Here, Cl- is the RA and PbO2 is the OA. The half-reactions are: Oxidation (half-reaction of Cl-): \(2\,Cl^{-} (\mathrm{aq}) \rightarrow Cl_{2} (\mathrm{g}) + 2\,e^{-}\) Reduction (half-reaction of PbO2): \(PbO_{2} (\mathrm{s}) + 4\,H^{+} (\mathrm{a}) + 2\,e^{-} \rightarrow Pb^{2+} (\mathrm{aq}) + 2\,H_{2}O (\mathrm{l})\) Now, balance the half-reactions in terms of the electrons and then add them together to yield the overall redox equation: \(PbO_{2} (\mathrm{s}) + 2\,Cl^{-} (\mathrm{aq}) + 4\,H^{+} (\mathrm{aq}) \rightarrow Pb^{2+} (\mathrm{aq}) + 2\,H_{2}O (\mathrm{l}) + Cl_{2} (\mathrm{g})\)

The Nernst equation is \(E=E^{0}-\frac{0.0592}{n}\log \frac{[product]}{[reactant]}\) where \(E\) is the cell potential, \(E^{0}\) is the standard cell potential, \(n\) is the number of electrons transferred, and the last term represents the ratio of product to reactant concentrations at non-standard conditions. For (a) \(\left[\mathrm{H}^{+}\right]=6.0 \mathrm{M}\): Substitute the given values into the Nernst equation to determine \(E\). Remember that if \(E > 0\), the reaction is spontaneous, and if \(E < 0\), it is non-spontaneous. Here is the same process for (b) \(\left[\mathrm{H}^{+}\right]=1.2 \mathrm{M}\) and (c) \(\mathrm{pH}=4.25\) (where \(\mathrm{pH} = -\log[\mathrm{H}^+]\), meaning \(\left[\mathrm{H}^{+}\right] = 10^{-4.25}\)).