A voltaic cell that uses the reaction $$ \operatorname{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

First, we need to determine the two half-cell reactions. One of them involves Cd and the other one involves Pd. From the overall reaction, we can break them into two half-reactions: 1. Pd-containing half-reaction: \[ \mathrm{PdCl}_{4}^{2-}(a q) \longrightarrow\mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+2 e^{-} \] Here, Pd is being reduced as it gains 2 electrons. 2. Cd-containing half-reaction: \[ \mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+2 e^{-} \] In this half-reaction, Cd is being oxidized as it loses 2 electrons.

From Appendix E, we can find the standard reduction potential (E°) for the Cd-containing reaction as: \[ \mathrm{Cd^{2+}(a q)+2 e^{-}} \longrightarrow \mathrm{Cd}(s);\quad E_{\text { red }}^{\circ} = -0.40\ \text{V} \] We are also given the standard cell potential (E°cell) for the voltaic cell: \[ E_{\text { cell }}^{\circ} = +1.03\ \text{V} \] Using the Nernst equation for standard cell potential, \[ E_{\text { cell }}^{\circ} = E_{\text { cathode }}^{\circ} -E_{\text { anode }}^{\circ} \] Since the standard cell potential is positive, the Pd-containing half-reaction must have the higher reduction potential and therefore must be the cathode. The Cd-containing half-reaction is then the anode. Therefore, we can write the Nernst equation as: \[ E_{\text { cell }}^{\circ} = E_{\text {Pd-red }}^{\circ} - (-0.40\ \text{V}) \] Now solve for the standard reduction potential (E°) for the Pd-containing reaction: \[ E_{\text {Pd-red }}^{\circ} = E_{\text { cell }}^{\circ} +0.40\ \text{V} = 1.03\ \text{V} + 0.40\ \text{V} = 1.43\ \text{V} \]

To sketch the voltaic cell, you can follow these steps: 1. Draw two containers, one for the anode half-cell and the other for the cathode half-cell. 2. In the anode half-cell (left side), place a Cd electrode, and label it "Anode" along with the Cd half-reaction: \[ \mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+2 e^{-} \] 3. In the cathode half-cell (right side), place a Pd electrode, and label it "Cathode" along with the Pd half-reaction: \[ \mathrm{PdCl}_{4}^{2-}(a q) \longrightarrow\mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+2 e^{-} \] 4. Connect the two electrodes with an external wire to allow electron flow from the anode (Cd) to the cathode (Pd), and place the voltmeter across the wire. (Don't forget to draw an arrow showing the electron flow direction from anode to cathode). 5. Add a salt bridge between the two cells to ensure electrical neutrality. The salt bridge contains ions that can flow between the two half-cells, balancing the charges. In the end, your sketched voltaic cell should show two half-cells with their respective electrodes, the external wire with an arrow indicating the electron flow, and the salt bridge.