A voltaic cell is made by combining the half-reactions:
Sn²⁺ + 2 e^-→Sn E ° = +0.15 V Ag⁺ + e^-→Ag E ° = +0.80 V Calculate the voltage of the cell at 298 K if [Sn²⁺] = 0.15 M and [Ag⁺] = 1.7 M.

Question

Quizplus · Accepted Answer

The voltage of the cell can be calculated using the equation: Ecell = Ered, cathode - Ered, anode where Ered, cathode is the reduction potential of the cathode and Ered, anode is the reduction potential of the anode. The given half-reactions are: Sn²⁺ + 2 e^- → Sn E ° = +0.15 V Ag⁺ + e^- → Ag E ° = +0.80 V The cathode is Ag (reduction half-reaction) and the anode is Sn (oxidation half-reaction). Therefore, the reduction potential of the cathode (Ered, cathode) is +0.80 V and the reduction potential of the anode (Ered, anode) is -0.15 V (as it is an oxidation half-reaction, its potential is the negative of the reduction potential). The concentration of the reactants and products also needs to be taken into consideration. The reduction half-reaction involves Ag+ and the oxidation half-reaction involves Sn2+. Therefore, the concentration of Ag+ is 1.7 M and the concentration of Sn2+ is 0.15 M. Substituting the values in the equation for Ecell, we get: Ecell = +0.80 V - (-0.15 V) - (0.0592 V / 2) log (1.7 / 0.15) (where 0.0592 V is the value of the Faraday constant F at 298 K) Simplifying the equation gives: Ecell = +0.65 V Therefore, the answer is B.

A Voltaic Cell Is Made by Combining the Half-Reactions