A galvanic cell is constructed as follows. One halfcell consists of a platinum wire immersed in a solution containing \(1.0 M \mathrm{Sn}^{2+}\) and \(1.0 M \mathrm{Sn}^{4+} ;\) the other half-cell has a thallium rod immersed in a solution of \(1.0 M \mathrm{TI}^{+}\). (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (c) What is the cell voltage if the \(\mathrm{TI}^{+}\) concentration is increased tenfold? \(\left(E_{\mathrm{T} 1^{+} / \mathrm{T} 1}^{\circ}=-0.34 \mathrm{~V} .\right)\)

The reactions occurring in the two half-cells are represented as:\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \] (1)\[ \text{Tl}^{+} + e^- \rightarrow \text{Tl} \] (2)In a galvanic cell, the oxidation reaction (loss of electrons) takes place at the anode and the reduction (gain of electrons) at the cathode.Ti1+ is being reduced to Ti. Therefore, it acts as the cathode (which is the right half cell by convention). Sn4+ is being reduced to Sn2+ but at the same time, it is also acting as an oxidising agent for Ti, therefore it acts as the anode (which is the left half cell by convention). The overall reaction for the cell is the sum of the anode and the cathode reactions:\[ \text{Sn}^{4+} + \text{Tl} \rightarrow \text{Sn}^{2+} + \text{Tl}^{+} \] (3)

To calculate the equilibrium constant (K), we need the reduction potentials for the two half reactions, given by \(E^\circ\). However, in the problem, only the reduction potential for Thallium is given. Reduction of Tin is not provided, and we have to infer it from the fact that the cell has zero voltage when equilibrium is established. We can use the equation for the standard cell potential: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0\] Substituting for the known reduction potentials gives: \(0 = -0.34 \, \text{V} - E^\circ_{\text{Sn4+/Sn2+}}\)This equation can be solved to give the unknown reduction potential as:\[ E^\circ_{\text{Sn4+/Sn2+}} = -0.34 \, \text{V} \]Then use the Nernst equation in the form that deals with the standard cell potential and equilibrium constant: \[E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K \]When rearranged, gives:\[ K = e^{nFE^\circ_{\text{cell}}/RT} \]Given that n = 1 (one electron is exchanged in the overall reaction), R = 8.314 (\text{J/mol} \, \text{K}), and T = 298 K, we can substitute these values, along with the given standard cell potential of 0 V to calculate K:\[ K = e^{0} = 1 \]

The cell voltage with a tenfold increase in the Tl+ ion concentration can be calculated using the Nernst equation:\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]Here, the reaction quotient Q is given by [Tl+][Sn2+]/[Sn4+]. Or in this case, since the [Tl+] and [Sn2+] increases tenfold, Q=[Tl+]\times 10/[Sn2+], and the Nernst equation simplifies to:\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln 10 \]Substituting known values gives the cell voltage:\[E_{\text{cell}} = 0 \, \text{V} - \frac{8.314 \, \text{J/mol K} \times 298 \, \text{K}}{1 \times 96485 \, \text{C/mol}} \ln 10 \approx -0.03 \, \text{V}\]