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

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To find the ratio of [Fe²⁺]/[Cd²⁺], we need to use the Nernst equation which states that: E_cell = E°cell - (RT/nF) ln(Q) where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, and F is the Faraday constant. At equilibrium, Q = [Fe²⁺]/[Cd²⁺] So, we can rearrange the Nernst equation to solve for this ratio: ln([Fe²⁺]/[Cd²⁺]) = (E°Fe - E°Cd)/(0.0257 V) Plugging in the given values, we get: ln([Fe²⁺]/[Cd²⁺]) = (-0.44 V + 0.40 V)/(0.0257 V) = -1.56 Taking the antilog of both sides: [Fe²⁺]/[Cd²⁺] = 10^(-1.56) = 2 × 10^(-2) = 2 × 10^(2-4) = 2 × 10^(-2+4) = 2 × 10^2 Therefore, the ratio [Fe²⁺]/[Cd²⁺] is 2 × 10^-2 or 2 × 10^2 in scientific notation. The closest answer choice is A.

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

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