(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{Cl}_{2}, \mathrm{O}_{2} .\) (b) Arrange the fol- lowing in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Al}\).

The first step is to find the reduction potentials for each chemical species. You can find this information in a standard reduction potential table, which is usually available in chemistry textbooks or online. The standard reduction potentials at 25°C in acidic solution are as follows: - \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\) : \(E^\circ = 1.33\,V\) - \(\mathrm{H}_{2} \mathrm{O}_{2} + 2\mathrm{H}^+ + 2e^- \rightarrow 2\mathrm{H}_2\mathrm{O}\) : \(E^\circ = 1.77\,V\) - \(\mathrm{ Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\) : \(E^\circ = 0.34\,V\) - \(\mathrm{ Cl}_{2} + 2e^- \rightarrow 2\mathrm{ Cl}^-\) : \(E^\circ = 1.36\,V\) - \(\mathrm{O}_{2} + 4\mathrm{H}^+ + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O}\) : \(E^\circ = 1.23\,V\) And for the reducing agents: - \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^-\) : \(E^\circ = -0.76\,V\) - \(\mathrm{I}^{-} \rightarrow \mathrm{I}_2 + 2e^-\) : \(E^\circ = 0.54\,V\) - \(\mathrm{ Sn}^{2+} \rightarrow \mathrm{Sn}^{4+} + 2e^-\) : \(E^\circ = 0.15\,V\) - \(\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} + 2e^-\) : \(E^\circ = 1.77\,V\) (It is both oxidizing and reducing agent) - \(\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3e^-\) : \(E^\circ = -1.66\,V\)

Now we need to arrange the oxidizing agents in increasing order of reduction potential value. As we recall, a higher reduction potential means it is a stronger oxidizing agent. So, we can arrange them as follows: \(\mathrm{Cu}^{2+} < \mathrm{O}_{2} < \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} < \mathrm{Cl}_{2} < \mathrm{H}_{2} \mathrm{O}_{2}\)

Lastly, we need to arrange the reducing agents in increasing order of reduction potential value. Remember, a lower reduction potential means it is a stronger reducing agent. So, we can arrange them as follows: \(\mathrm{Al} < \mathrm{Zn} < \mathrm{Sn}^{2+} < \mathrm{I}^{-} < \mathrm{H}_{2} \mathrm{O}_{2}\) Therefore, the final order is: (a) Oxidizing agents: \(\mathrm{Cu}^{2+} < \mathrm{O}_{2} < \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} < \mathrm{Cl}_{2} < \mathrm{H}_{2} \mathrm{O}_{2}\) (b) Reducing agents: \(\mathrm{Al} < \mathrm{Zn} < \mathrm{Sn}^{2+} < \mathrm{I}^{-} < \mathrm{H}_{2} \mathrm{O}_{2}\)