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 atm. a. $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}$ $\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}$ b. $\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}$ $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}$

Comparing the given \(\mathscr{E}^{\circ}\) of both half-reactions, 1.78 V > 0.68 V Therefore, the first half-reaction will function as the cathode and the second as the anode.

Now, we'll combine the two half-reactions to find the overall cell reaction: \(\mathrm{H}_{2} \mathrm{O}_{2} + 2 \mathrm{H}^{+} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (cathode) \(\mathrm{O}_{2} + 2 \mathrm{H}^{+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}\) (anode) Adding these reactions, we get: \(\mathrm{O}_{2} + 2 \mathrm{H}^{+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} + \mathrm{H}_{2} \mathrm{O}_{2} + 2 \mathrm{H}^{+} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) Which can be simplified to: \(\mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\)

To determine \(\mathscr{E}^{\circ}\) for the cell, subtract the anode potential from the cathode potential: \(\mathscr{E}^{\circ}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cathode}} - \mathscr{E}^{\circ}_{\text{anode}}\) \(\mathscr{E}^{\circ}_{\text{cell}} = 1.78 - 0.68 = 1.10\) V For part b, we will perform the same calculations:

Comparing the given \(\mathscr{E}^{\circ}\) of both half-reactions, -1.18 V < -0.036 V Therefore, the first half-reaction will function as the anode and the second as the cathode.

Now, we'll combine the two half-reactions to find the overall cell reaction: \(\mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\) (anode) \(\mathrm{Fe}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}\) (cathode) Since the number of electrons differ, we must multiply the first equation by 3 and the second by 2 such that the transferred electrons are the same: Anode: \(3(\mathrm{Mn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}) \Rightarrow 3\mathrm{Mn}^{2+} + 6 \mathrm{e}^{-} \rightarrow 3\mathrm{Mn}\) Cathode: \(2(\mathrm{Fe}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}) \Rightarrow 2\mathrm{Fe}^{3+} + 6 \mathrm{e}^{-} \rightarrow 2\mathrm{Fe}\) Adding these reactions, we get: \(3\mathrm{Mn}^{2+} + 6 \mathrm{e}^{-} \rightarrow 3\mathrm{Mn} + 2\mathrm{Fe}^{3+} + 6 \mathrm{e}^{-} \rightarrow 2\mathrm{Fe}\) Which can be simplified to: \(3\mathrm{Mn}^{2+} + 2\mathrm{Fe}^{3+} \rightarrow 3\mathrm{Mn} + 2\mathrm{Fe}\)

To determine \(\mathscr{E}^{\circ}\) for the cell, subtract the anode potential from the cathode potential: \(\mathscr{E}^{\circ}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cathode}} - \mathscr{E}^{\circ}_{\text{anode}}\) \(\mathscr{E}^{\circ}_{\text{cell}} = -0.036 - (-1.18) = 1.14\) V