Final answer:
The ratio of [Fe²⁺] to [Cd²⁺] in a galvanic cell with E = 0.000 V and E° = 0.0400 V at 298 K is determined using the Nernst equation, which involves the reaction quotient Q and the number of electrons transferred in the half-reactions.
Step-by-step explanation:
The student is asking about the ratio of concentrations [Fe²⁺] to [Cd²⁺] in a galvanic cell at 298 K where the cell potential (E) is 0.000 V. To find this ratio, we use the Nernst equation which at 298 K simplifies to E = E° - (0.0592/n) log Q, where n is the number of electrons transferred in the half-reaction, E° is the standard cell potential, and Q is the reaction quotient. In this case, the standard cell potential (E°) is provided as 0.0400 V, and n is 2 because each reaction involves the transfer of 2 electrons. The reaction quotient, Q, is defined as the ratio of the concentrations of the products over the reactants, considering the coefficients of the balanced equation. Since E is given as 0.000 V, we can rearrange the Nernst equation to solve for Q: 0.000 = 0.0400 - (0.0592/2) log Q and therefore log Q = 0.0400 / (0.0592/2), giving us the value for Q, and hence the ratio [Fe²⁺]/[Cd²⁺].