Question

A \(1 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A \(1 M\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass and which loses 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) Anode: Sn electrode, Cathode: Cu electrode b) Sn electrode loses mass, Cu electrode gains mass c) Overall cell reaction: \( \mathrm{Sn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Cu} \) d) Cell emf under standard conditions: \(E_\mathrm{cell}^\circ = 0.473\,\mathrm{V}\)

Step-by-step solution

Identify the reduction reactions

First, we should write down the reduction half-reactions for both metals: \[ \mathrm{Cu}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Cu} \hspace{1cm} (E^\circ_\mathrm{Cu^{2+}/Cu} = +0.337\ \mathrm{V}) \] \[ \mathrm{Sn}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Sn} \hspace{1cm} (E^\circ_\mathrm{Sn^{2+}/Sn} = -0.136\ \mathrm{V}) \]

Identify the anode and cathode

In an electrochemical cell, the anode is the electrode where the oxidation occurs, and the cathode is where the reduction occurs. To determine which is the anode and which is the cathode, we compare the reduction potentials of the two half-reactions. The electrode with a higher reduction potential will be the cathode, and the one with a lower reduction potential will be the anode. In this case, \(E^\circ_\mathrm{Cu^{2+}/Cu}\) is greater than \(E^\circ_\mathrm{Sn^{2+}/Sn}\), so: - Anode: \(\mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2 \mathrm{e}^-\) - Cathode: \(\mathrm{Cu}^{2+} + 2 \mathrm{e}^- \rightarrow \mathrm{Cu}\)

Electrode mass changes

The anode is the site of oxidation, where the metal loses electrons and goes into solution as a positive ion. Therefore, the mass of the anode will decrease. The cathode is where reduction occurs and metal cations gain electrons and plate onto the electrode as a solid. Therefore, the mass of the cathode will increase. In this cell, Sn is the anode and Cu is the cathode. So, the Sn electrode loses mass, and the Cu electrode gains mass.

Overall cell reaction

To find the overall cell reaction, we combine the half-reactions (paying attention to the number of electrons): \[ \mathrm{Sn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Cu} \]

Calculate the cell emf

We can calculate the cell emf under standard conditions using the Nernst equation: \[ E_\mathrm{cell}^\circ = E_\mathrm{cathode}^\circ - E_\mathrm{anode}^\circ \] Using the reduction potential values from Step 1, we get: \[ E_\mathrm{cell}^\circ = (+0.337\,\mathrm{V}) - (-0.136\,\mathrm{V}) = 0.473\,\mathrm{V} \] Now we have answered all the questions: a) Anode: Sn electrode, Cathode: Cu electrode b) Sn electrode loses mass, Cu electrode gains mass c) Overall cell reaction: \( \mathrm{Sn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{Cu} \) d) Cell emf under standard conditions: \(E_\mathrm{cell}^\circ = 0.473\,\mathrm{V}\)