Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

First, let's figure out the balanced reaction for cell a. We have: Oxidation: \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) We need to balance the number of electrons in both half-reactions so that the total number of electrons in oxidation and reduction are equal. To do this, we can multiply the oxidation half-reaction by 3.

Now our balanced half-reactions are: Oxidation: \(3\mathrm{(H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O})\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) Now we can add the half-reactions together to get the balanced cell reaction: \(3(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 3(2 \mathrm{H}_{2} \mathrm{O}) + 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) Simplify the equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}_{2} \mathrm{O}_{2} + 20\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 13\mathrm{H}_{2} \mathrm{O}\)

Using Table 18.1, we can find the standard reduction potentials for the half-reactions: Oxidation (reverse of the original reaction): \(-\mathscr{E}^{\circ}(\mathrm{H}_{2} \mathrm{O}_{2}) = -1.77 \,\text{V}\) Reduction: \(\mathscr{E}^{\circ}(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}) = 1.33 \,\text{V}\) Now we can calculate the total cell potential: \(\mathscr{E}^{\circ}_{total}=\mathscr{E}^{\circ}_{reduction} + (-\mathscr{E}^{\circ}_{oxidation})\) \(\mathscr{E}^{\circ}_{total}= 1.33 \,\text{V} - (-1.77 \,\text{V})\) \(\mathscr{E}^{\circ}_{total}= 3.10 \,\text{V}\) _Cell b_

Similarly, let's balance the half-reactions for cell b: Oxidation: \(\mathrm{H}_{2} \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Reduction: \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}\) Both half-reactions are already balanced in terms of electrons.

Now we can add the half-reactions together to get the balanced cell reaction: \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} + \mathrm{Al}\) Simplify the equation: \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \longrightarrow 2 \mathrm{H}^{+} + \mathrm{Al}\)

Using Table 18.1, we can find the standard reduction potentials for the half-reactions: Oxidation (reverse of the original reaction): \(-\mathscr{E}^{\circ}(\mathrm{H}_{2}) = 0 \,\text{V}\) Reduction: \(\mathscr{E}^{\circ}(\mathrm{Al}^{3+}) = -1.66 \,\text{V}\) Now we can calculate the total cell potential: \(\mathscr{E}^{\circ}_{total}=-\mathscr{E}^{\circ}_{oxidation} + \mathscr{E}^{\circ}_{reduction}\) \(\mathscr{E}^{\circ}_{total}= 0 \,\text{V} - (-1.66 \,\text{V})\) \(\mathscr{E}^{\circ}_{total}= 1.66 \,\text{V}\) So, the balanced cell equations and their corresponding standard cell potentials are: a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}_{2} \mathrm{O}_{2} + 20\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 13\mathrm{H}_{2} \mathrm{O}\) with \(\mathscr{E}^{\circ}= 3.10 \,\text{V}\) b. \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \rightarrow 2 \mathrm{H}^{+} + \mathrm{Al}\) with \(\mathscr{E}^{\circ}= 1.66 \,\text{V}\)