Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm} .\) \(\begin{array}{ll}\text { a. } \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathscr{6}^{\circ}=1.78 \mathrm{~V} \\ \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} & \mathscr{6}^{\circ}=0.68 \mathrm{~V}\end{array}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\) \(\mathscr{6}^{\circ}=-1.18 \mathrm{~V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{6}^{\circ}=-0.036 \mathrm{~V}\)

In the first galvanic cell, the anode (oxidation) half-reaction is \(\mathrm{O}_{2} + 2\mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{O}_{2}\), and the cathode (reduction) half-reaction is \(\mathrm{H}_{2}\mathrm{O}_{2} + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{H}_{2}\mathrm{O}\). The overall balanced equation is \(\mathrm{O}_{2} + 2\mathrm{H}^{+} \rightarrow 2\mathrm{H}_{2}\mathrm{O}\), with \(\mathscr{E}^{\circ}_{cell} = 1.1\mathrm{V}\). In the second galvanic cell, the anode half-reaction is \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\), and the cathode half-reaction is \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}\). The overall balanced equation is \(3\mathrm{Mn}^{2+} + 2\mathrm{Fe}^{3+} \rightarrow 3\mathrm{Mn} + 2\mathrm{Fe}\), with \(\mathscr{E}^{\circ}_{cell} = 1.144\mathrm{V}\). In both cells, electrons flow from the anode to the cathode, and ions migrate through the salt bridge towards their respective electrode.