Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C: Sn(s) ∣ Sn²⁺(aq, 0.100 mol L^-1) Ag⁺(aq, 0.200 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: Sn(s) ∣ Sn²⁺(aq, 0.100 mol L^-1)

Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C: Sn(s) ∣ Sn<sup>2+</sup>(aq, 0.100 mol L<sup>-1</sup>) Ag<sup>+</sup>(aq, 0.200 mol L<sup>-1</sup>) ∣ Ag(s) E°(Sn<sup>2+</sup>/Sn) = -0.14 V and E°(Ag<sup>+</sup>/Ag) = +0.80 V A) -2.68 V B) -0.03 V C) +0.95 V D) +0.93 V E) +2.10 V

Ag⁺(aq, 0.200 mol L^-1) ∣ Ag(s)
E°(Sn²⁺/Sn) = -0.14 V and E°(Ag⁺/Ag) = +0.80 V

Quizplus · Accepted Answer

The cell potential of an electrochemical cell can be calculated using the formula: Ecell = E°cell - (RT/nF) * ln(Q) where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the balanced chemical equation, F is Faraday's constant, and Q is the reaction quotient. In this case, the balanced chemical equation for the cell reaction is: Sn(s) + 2Ag⁺(aq) -> Sn²⁺(aq) + 2Ag(s) The number of electrons transferred is 2, so n = 2. The temperature is 25°C, which is 298 K. The gas constant is 8.314 J/K*mol and Faraday's constant is 96485 C/mol. The reaction quotient Q can be calculated using the concentrations of the species involved: Q = [Sn²⁺(aq)] / ([Ag⁺(aq)]^2) Q = (0.100 mol/L) / (0.200 mol/L)^2 Q = 2.50 Substituting the values into the formula, we get: Ecell = E°cell - (RT/nF) * ln(Q) Ecell = (0.80 V) - [(8.314 J/K*mol) * (298 K) / (2 * 96485 C/mol)] * ln(2.5) Ecell = 0.93 V Therefore, the cell potential is +0.93 V. The best choice is D.

Calculate the Cell Potential for the Following Reaction That Takes