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

In the given reaction, \[ \mathrm{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) \] We can break this down into the following half-reactions: 1. \(\mathrm{Cd}(s) \longrightarrow \mathrm{Cd}^{2+}(a q)+ 2 e^{-}\) (oxidation) 2. \( \mathrm{PdCl}_{4}^{2-}(a q)+ 2 e^{-} \longrightarrow \mathrm{Pd}(s)+ 4 \mathrm{Cl}^{-}(a q)\) (reduction)

The given standard cell potential is \(+1.03 \mathrm{~V}\). To determine the reduction potential for the Pd half-reaction, we will first find the reduction potential for the oxidation half-reaction (Cd) from Appendix E - Standard Reduction Potentials. From Appendix E, the standard reduction potential for the Cd half-reaction is: \[ \mathrm{Cd}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Cd}(s); E_{\mathrm{red}}^{\circ} = -0.40 \mathrm{~V} \] Since the Cd is actually being oxidized, we will use the opposite of the reduction potential to obtain its oxidation potential: \[ E_{\mathrm{ox}}^{\circ}(\mathrm{Cd}) = -E_{\mathrm{red}}^{\circ}(\mathrm{Cd}) = +0.40 \mathrm{~V} \] Now, using the standard cell potential given, we can determine the reduction potential for the Pd half-reaction, \(E_{\mathrm{red}}^{\circ}(\mathrm{Pd})\): \(E_{\mathrm{cell}}^{\circ}=E_{\mathrm{red}}^{\circ}(\mathrm{Pd})-E_{\mathrm{ox}}^{\circ}(\mathrm{Cd})\) Now plug the values and solve: \(1.03\mathrm{~V}=E_{\mathrm{red}}^{\circ}(\mathrm{Pd}) - 0.40 \mathrm{~V}\) \(E_{\mathrm{red}}^{\circ}(\mathrm{Pd})=1.43 \mathrm{~V}\) So, the reduction potential for the reaction involving Pd is \(1.43 \mathrm{~V}\).

To sketch the voltaic cell, we need to label the anode and the cathode, as well as indicate the direction of electron flow: 1. Anode (oxidation occurs): Solid Cd 2. Cathode (reduction occurs): PdCl2 (aq) The electron flow will be from the anode (Cd) to the cathode (Pd). In our sketch, we will have two containers (half-cells). In one container, we will have solid Cd as the anode, immersed in a solution containing Cd2+ ions. In the other container, we will have Pd electrode, immersed in a solution containing PdCl42- ions. A salt bridge will be connecting the two containers to balance the charges, while a wire connects the electrodes, indicating the direction of electron flow from Cd to Pd.