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. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

The two half-reactions are: 1. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) (Reduction half-reaction) 2. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (Oxidation half-reaction)

We need to multiply each half-reaction to make the number of electrons equal in both. Multiply the second reaction with 3, so we'll have 6 electrons in each reaction: 1. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) 2. \(3(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}) \Rightarrow 3\mathrm{H}_{2}\mathrm{O}_{2}+6\mathrm{H}^{+}+6\mathrm{e}^{-}\rightarrow 6\mathrm{H}_{2} \mathrm{O}\)

Add the two balanced half-reactions to get the overall cell reaction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-}+3\mathrm{H}_{2}\mathrm{O}_{2}+6\mathrm{H}^{+}+6\mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+6\mathrm{H}_{2} \mathrm{O}\) Simplify the equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3\mathrm{H}_{2}\mathrm{O}_{2}+20\mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+13\mathrm{H}_{2}\mathrm{O}\)

The standard cell potential \(\mathscr{E}^{\circ}\) is the difference between the standard reduction potentials of the two half-reactions. Table 18.1 values: \(\mathscr{E}^{\circ}_\text{Cr}=1.33\,\text{V}\) (for the first half-reaction) \(\mathscr{E}^{\circ}_\text{H2O2}=1.77\,\text{V}\) (for the second half-reaction) Since \(\mathrm{H}_{2} \mathrm{O}_{2}\) is the oxidizing agent (it's being reduced), we'll have: \(\mathscr{E}^{\circ}=\mathscr{E}^{\circ}_\text{Cr}-\mathscr{E}^{\circ}_\text{H2O2}=1.33\,\text{V}-1.77\,\text{V}=-0.44\,\text{V}\) For the second galvanic cell:

The two half-reactions are: 1. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) (Reduction half-reaction) 2. \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\) (Oxidation half-reaction)

Multiply the first reaction by 3 to equalize the number of electrons in each half-reaction: 1. \(3(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2})\Rightarrow 6\mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}\) 2. \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Add the two balanced half-reactions to get the overall cell reaction: \(6\mathrm{H}^{+}+6 \mathrm{e}^{-}+\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}+\mathrm{Al}\)

The standard cell potential \(\mathscr{E}^{\circ}\) is the difference between the standard reduction potentials of the two half-reactions. Table 18.1 values: \(\mathscr{E}^\circ_\text{H}=0\,\text{V}\) (for the first half-reaction) \(\mathscr{E}^\circ_\text{Al}=-1.66\,\text{V}\) (for the second half-reaction) Since \(\mathrm{H}^{+}\) is the oxidizing agent (it's being reduced), we'll have: \(\mathscr{E}^{\circ}=\mathscr{E}^{\circ}_\text{H}-\mathscr{E}^{\circ}_\text{Al}=0\,\text{V}-(-1.66\,\text{V})=1.66\,\text{V}\) In summary for both galvanic cells, the balanced cell equations and standard cell potentials are: a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3\mathrm{H}_{2}\mathrm{O}_{2}+20\mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+13\mathrm{H}_{2}\mathrm{O}\), \(\mathscr{E}^{\circ}=-0.44\,\text{V}\) b. \(6\mathrm{H}^{+}+\mathrm{Al}^{3+}+9\mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}+\mathrm{Al}\), \(\mathscr{E}^{\circ}=1.66\,\text{V}\)