To calculate the cell potential for the reaction, we first need to determine the oxidation and reduction half-reactions. The given reaction is:Hg2Cl2 s + 2 Cl^- aq 2 Hg l + 2 Cl2 g The given standard reduction potentials are:1 Hg2Cl2 s + 2 e^- 2 Hg l + 2 Cl^- aq E = 0.268 V2 Cl2 g + 2 e^- 2 Cl^- aq E = 1.36 VIn the overall reaction, Hg2Cl2 is being reduced to Hg, so we can use the first half-reaction as is. However, Cl^- is being oxidized to Cl2, so we need to reverse the second half-reaction:2 Cl^- aq Cl2 g + 2 e^- E = -1.36 VNow, we can add the two half-reactions together:Hg2Cl2 s + 2 e^- + 2 Cl^- aq 2 Hg l + 2 Cl^- aq + Cl2 g + 2 e^-Simplifying the reaction, we get:Hg2Cl2 s + 2 Cl^- aq 2 Hg l + Cl2 g Now, we can calculate the cell potential by adding the standard reduction potentials of the two half-reactions:E cell = E reduction + E oxidation E cell = 0.268 V + -1.36 V E cell = -1.092 VThe cell potential for the reaction is -1.092 V.