Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C:
$\quad$ $\quad$ $\quad$ Sn(s) | Sn²⁺(aq,0.010 mol L^-1) $\|$ Ag⁺(aq,1.00 mol L^-1) | Ag(s)
E°(Sn²⁺/Sn) = -0.14 V and E°(Ag⁺/Ag) = +0.80 V

Question

Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C:

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Sn(s) | Sn²⁺(aq,0.010 mol L^-1)

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Ag⁺(aq,1.00 mol L^-1) | Ag(s)
E°(Sn²⁺/Sn) = -0.14 V and E°(Ag⁺/Ag) = +0.80 V

Quizplus · Accepted Answer

The cell potential can be calculated using the formula: Cell potential (Ecell) = E°(reduction at cathode) - E°(oxidation at anode) E°(Sn²⁺/Sn) is the oxidation reaction, where Sn is being oxidized to Sn²⁺, so the sign of E° is flipped to represent the oxidation reaction: E°(oxidation) = +0.14 V E°(Ag⁺/Ag) is the reduction reaction, where Ag⁺ is being reduced to Ag, so the E° remains positive: E°(reduction) = +0.80 V Plugging these values into the formula: Ecell = (+0.80 V) - (+0.14 V) Ecell = +0.66 V But this value is for standard conditions (1 M concentration), not the given concentration of 0.010 M. Therefore, we need to use the Nernst equation to adjust for concentration: Ecell = E°cell - (0.0592 V/n) x log(Q) where n is the number of electrons transferred (in this case, 2), and Q is the reaction quotient, which can be calculated using the concentrations of the species involved in the reaction: Q = [Ag+] / [Sn²⁺] Q = (1.00 M) / (0.010 M) Q = 100 Plugging this into the Nernst equation: Ecell = +0.66 V - (0.0592 V/2) x log(100) Ecell = +1.00 V Therefore, the cell potential is +1.00 V, and the best choice is C.

Calculate the Cell Potential for the Following Reaction That Takes \quad

Calculate the Cell Potential for the Following Reaction That Takes $\quad$