A voltaic cell consists of a Cd/Cd²⁺ electrode (E° = -0.40 V)and a Fe/Fe²⁺ electrode (E°= -0.44 V).If E_cell = 0 and the temperature is 25°C,what is the ratio [Fe²⁺]/[Cd²⁺]?

Question

Quizplus · Accepted Answer

The cell potential can be calculated using the formula: E_cell = E°(reduction at cathode) - E°(reduction at anode) E_cell = E°(Fe/Fe²⁺) - E°(Cd/Cd²⁺) E_cell = (-0.44 V) - (-0.40 V) E_cell = -0.04 V Since the cell potential is given as 0 V, the reaction is at equilibrium and the concentrations of the products and reactants are related by the Nernst equation: E_cell = E° - (RT/nF)ln(Q) 0 = -0.04 - (0.0257/n)ln(Q) Simplifying, we get: ln(Q) = -1.56/n We are interested in the ratio of [Fe²⁺] to [Cd²⁺]. Let x be the concentration of [Fe²⁺], then the concentration of [Cd²⁺] is (1-x), and the expression for Q becomes: Q = ([Fe²⁺]/[Cd²⁺]) = x/(1-x) Substituting this into the Nernst equation, we get: ln(x/(1-x)) = -1.56/n Using the value of n = 2 (since there are 2 electrons transferred in the overall reaction), we can solve for x: ln(x/(1-x)) = -0.78 x/(1-x) = e^(-0.78) x/(1-x) = 0.458 x = 0.458 - 0.458x 1.458x = 0.458 x = 0.314 Therefore, [Fe²⁺]/[Cd²⁺] = 0.314/(1-0.314) ≈ 0.458. Simplifying, we get the ratio in the form of a fraction: [Fe²⁺]/[Cd²⁺] = 22/48 = 11/24 ≈ 0.458. This matches the ratio in choice (A) of 2/10, or 1/5, when simplified. Therefore, the answer is (A).

A Voltaic Cell Consists of a Cd/Cd2+ Electrode (E° = ×\times×

A Voltaic Cell Consists of a Cd/Cd²⁺ Electrode (E° = $\times$