(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: $\mathrm{MnO}_{4}^{-}(a q), \mathrm{O}_{3}(g), \mathrm{HSO}_{4}^{-}(a q), \mathrm{O}_{2}(g), \mathrm{HClO}(a q)$ (b) Arrange the following in order of increasing strength as reducing agents in basic solution: $\mathrm{Cr}(\mathrm{OH})_{3}(s), \mathrm{Fe}(s), \mathrm{Ca}(s),\( \)\mathrm{H}_{2}(g), \mathrm{Mn}(s)$

Rank the species by their standard half-cell potential from smallest to largest. Step 3: Order of increasing strength as oxidizing agents

Look up the standard half-cell potentials for each of the given species in basic solution: \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\), \(\mathrm{Fe}(s)\), \(\mathrm{Ca}(s)\), \(\mathrm{H}_{2}(g)\), and \(\mathrm{Mn}(s)\). These values can be found in standard reduction potential tables. Step 2: Compare the half-cell potentials

The order of increasing strength as reducing agents in basic solution corresponds to the decreasing half-cell potentials, as lower half-cell potential values indicate stronger reducing agents. List them in the order of their decreasing half-cell potentials. Answers: The standard half-cell potentials for the given species in the basic solution are as follows: \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\): \( -0.74 \, V \) \(\mathrm{Fe}(s)\): \( -0.45 \, V \) \(\mathrm{Ca}(s)\): \( -2.87 \, V \) \(\mathrm{H}_{2}(g)\): \( 0.00 \, V \) \(\mathrm{Mn}(s)\): \( -1.18 \, V \) Therefore, the order of increasing strength as reducing agents in the basic solution is: \(\mathrm{H}_{2}(g) < \mathrm{Fe}(s) < \mathrm{Cr}(\mathrm{OH})_{3}(s) < \mathrm{Mn}(s) < \mathrm{Ca}(s)\)