Given:
2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 1.00×10^{- 3} M, [U⁴⁺] = 0.200 M, [Cd²⁺] = 0.100 M?

Question

Given: 2 U³⁺ + Cd²⁺→Cd (s) + 2 U⁴⁺ with E ° = +0.207 V What is the potential of this cell reaction at 298 K when [U³⁺] = 1.00×10^{- 3} M, [U⁴⁺] = 0.200 M, [Cd²⁺] = 0.100 M?

Quizplus · Accepted Answer

The Nernst equation is used to calculate the cell potential under non-standard conditions: E = E° - (RT/nF) * ln(Q). Here, E° = +0.207 V, R = 8.314 J/mol·K, T = 298 K, n = 2, F = 96485 C/mol. The reaction quotient Q = [U^4+]^2 / ([U^3+]^2 * [Cd^2+]) = (0.200^2) / ((1.00×10^-3)^2 * 0.100). Plugging these values into the Nernst equation gives E ≈ +0.0415 V.

Given: 2 U3+ + Cd2+→Cd (S) + 2 U4+ with E

Given: 2 U³⁺ + Cd²⁺→Cd (S) + 2 U⁴⁺ with E