A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

We need to identify the two half-reactions occurring in the cell. In the first beaker, we have the cadmium half-reaction, which can be written as: \( Cd^{2+} + 2e^- \leftrightarrow Cd \) In the second beaker, we have the chlorine half-reaction: \( Cl_2 + 2e^- \leftrightarrow 2Cl^- \)

Recall that the anode is the electrode where oxidation occurs, whereas the cathode is the electrode where reduction occurs. We need to determine which of the two half-reactions is the oxidation half-reaction and which one is the reduction half-reaction. As a general rule, half-reactions with a more positive electrode potential (or reduction potential) will proceed as reductions in an electrochemical cell. For this problem, standard electrode potentials must be consulted (often available in a textbook or online references). We see that the electrode potential for the Cd half-reaction is \( -0.40V \) and for the Cl half-reaction is \( +1.36V \). Therefore, the Cl half-reaction will proceed as a reduction, making the platinum electrode the cathode, and the Cd half-reaction will proceed as an oxidation, making the cadmium electrode the anode.

Since the Cd half-reaction is an oxidation, the Cd metal will lose electrons and turn into \( Cd^{2+} \) ions. As a result, the Cd electrode will lose mass as the cell reaction proceeds.

Now that we have the anode and cathode half-reactions, we can combine them to find the overall cell reaction. At the anode (oxidation), the Cd half-reaction can be written as: \( Cd \rightarrow Cd^{2+} + 2e^- \) (oxidation half-reaction) At the cathode (reduction), the Cl half-reaction can be written as: \( Cl_2 + 2e^- \rightarrow 2Cl^- \) (reduction half-reaction) Adding the two half-reactions together gives the overall cell reaction: \( Cd + Cl_2 \rightarrow Cd^{2+} + 2Cl^- \)

To calculate the cell potential (emf), subtract the standard electrode potential of the anode from the standard electrode potential of the cathode: \( E_{cell} = E_{cathode} - E_{anode} \) \( E_{cell} = (+1.36V) - (-0.40V) \) \( E_{cell} = +1.76V \) Therefore, the emf generated by the cell under standard conditions is +1.76V.