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⁺] = 0.100 M, [U⁴⁺] = 2.00×10^{- 3} M, [Cd²⁺] = 0.200 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⁺] = 0.100 M, [U⁴⁺] = 2.00×10^{- 3} M, [Cd²⁺] = 0.200 M?

Quizplus · Accepted Answer

First, we need to write the balanced redox reaction and calculate the standard cell potential, E°. 2 U³⁺ + Cd²⁺ → Cd (s) + 2 U⁴⁺ E° = +0.207 V Using the Nernst equation, we can calculate the cell potential, E, at non-standard conditions: E = E° - (RT/nF) ln(Q) where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, n is the number of electrons transferred in the balanced redox reaction, F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient, which is the ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients. At 298K, we have: E = 0.207 V - (8.314 J/mol.K × 298 K / 2 mol electrons / 96485 C/mol) ln[(2.00×10^{- 3})² / (0.100)(0.200)] E = 0.207 V - 0.538 V E = +0.127 V Therefore, the cell potential at non-standard conditions is +0.127 V. The correct answer is D, +0.287 V.

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

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