A voltaic cell that uses the reaction $$ \mathrm{T}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tl}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) $$ has a measured standard cell potential of \(+1.19 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

In this voltaic cell, one half must be oxidation and the other must be reduction. The current overall cell reaction is: \[ \mathrm{Tl}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tl}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) \] Since the charges of Tl are decreasing from +3 to +1, this is a reduction reaction. On the other hand, the charges of Cr are increasing from +2 to +3, which is indicative of an oxidation reaction. So, we can write the two half-cell reactions as follows: Reduction: \[ \mathrm{Tl}^{3+}(a q) + 2e^- \longrightarrow \mathrm{Tl}^{+}(a q) \] Oxidation: \[ 2\mathrm{Cr}^{2+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 2e^- \]

The standard cell potential for the whole cell is given as +1.19 V. We can use the formula \(E°_{cell} = E°_{cathode} - E°_{anode}\) to find out the standard reduction potential for the reaction involving Pd. For the half-cell involving Tl, we can look up the standard reduction potential in Appendix E of the textbook. From Appendix E, we know that the standard reduction potential for the Cr half-cell is: \[ E°_{Cr^{3+}/Cr^{2+}} = -0.41 \mathrm{V} \] Now, plugging the known values into the formula: \[ 1.19 = E°_{Tl^{3+}/Tl^{+}} - (-0.41) \] Solving for the standard reduction potential for Tl half-cell \(E°_{Tl^{3+}/Tl^{+}}\): \[ E°_{Tl^{3+}/Tl^{+}} = 1.19 + 0.41 = 1.60 \mathrm{V} \]

In this voltaic cell, the anode will have the oxidation reaction, with chromium atoms giving up electrons, while the cathode will have the reduction reaction, with thallium atoms gaining electrons. 1. Label the anode with the oxidation half-reaction: \(2\mathrm{Cr}^{2+}(a q) \longrightarrow 2\mathrm{Cr}^{3+}(a q) + 2e^-\) 2. Label the cathode with the reduction half-reaction: \(\mathrm{Tl}^{3+}(a q) + 2e^- \longrightarrow \mathrm{Tl}^{+}(a q)\) 3. Show the electron flow with an arrow pointing from the anode to the cathode, as electrons flow from the oxidation half-cell to the reduction half-cell. The completed sketch should have labeled anode and cathode, showing half-cell reactions in each, with an arrow indicating electron flow from anode (Cr) to cathode (Tl).