Question

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 \(\mathrm{NaCl}\) 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?

Short answer

(a) The anode is the Cd electrode, and the cathode is the Pt electrode. (b) The Cd electrode loses mass. (c) The overall cell reaction is \(Cd + Cl_{2(g)} \rightarrow Cd^{2+} + 2Cl^-\). (d) The standard emf generated by the cell is 1.76 V.

Step-by-step solution

Identify the Half-cell Reactions

First, we need to identify the half-cell reactions occurring at each electrode. In the Cd(NO3)2 half-cell, the possible reaction is the reduction/oxidation of cadmium metal: \[Cd^{2+} + 2e^- \leftrightarrow Cd\] For the other half-cell, we have NaCl solution and Cl2 gas, which suggests the possible reaction is the reduction/oxidation of chlorine: \[2Cl^- \leftrightarrow Cl_{2(g)} + 2e^-\]

Determine the Anode and Cathode by Standard Reduction Potentials

Now, we need to look up the Standard reduction potentials (E°) of both half-cell reactions. The higher potential species will undergo reduction (be the cathode), and the lower potential species will undergo oxidation (be the anode). From the standard reduction table, we can find: \[E°_{Cd^{2+}/Cd} = -0.40 V\] \[E°_{Cl_{2}/Cl^-} = 1.36 V\] Since E°(Cl2/Cl-) is higher than E°(Cd2+/Cd), Cl2 will be reduced at the platinum electrode, making it the cathode. On the other hand, Cd will be oxidized at the cadmium electrode, making it the anode.

Overall Cell Reaction and Mass Change of the Cd Electrode

To find the overall cell reaction, we add the half-cell reactions together, making sure to balance the electrons. We can also determine whether the Cd electrode gains or loses mass in the process. Oxidation (at the anode): \[Cd \rightarrow Cd^{2+} + 2e^-\] Reduction (at the cathode): \[Cl_{2(g)} + 2e^- \rightarrow 2Cl^-\] Overall cell reaction: \[Cd + Cl_{2(g)} \rightarrow Cd^{2+} + 2Cl^-\] As the cell reaction proceeds, the Cd electrode loses mass because it is oxidized from Cd to Cd2+.

Calculate the Standard EMF

The standard electrochemical potential (known as emf or E°cell) can be calculated as the difference between reduction potential at the cathode (E°red) and reduction potential at the anode (E°oxi): \[E°_{cell} = E°_{red} - E°_{oxi}\] We already found the standard reduction potentials: \[E°_{Cd^{2+}/Cd} = -0.40 V\] \[E°_{Cl_{2}/Cl^-} = 1.36 V\] So the emf generated by the cell under standard conditions is: \[E°_{cell} = 1.36 V - (-0.40 V) = 1.76 V\] In summary: (a) Anode: Cd electrode; Cathode: Pt electrode (b) Cd electrode loses mass (c) Overall cell reaction: \(Cd + Cl_{2(g)} \rightarrow Cd^{2+} + 2Cl^-\) (d) Standard emf: 1.76 V