Consider the electrochemical cell described by the following reaction at 310 K. The measured cell potential is 0.242 V. What is the concentration of iodide ion (in M) if there are 1.40 M of cadmium ions in the cell? 2CuI(s) + Cd(s) → Cd2+ + 2I– + 2Cu(s) (Use R = 8.314 J/mol-K; F = 96485 C/mol) Additional data: E° CuI/I-/Cu = –0.17 V E° Cd2+/Cd = –0.403 V
Consider the electrochemical cell described by the following reaction at 310 K. The measured cell potential is 0.242 V. What is the concentration of iodide ion (in M) if there are 1.40 M of cadmium ions in the cell? 2CuI(s) + Cd(s) → Cd2+ + 2I– + 2Cu(s) (Use R = 8.314 J/mol-K; F = 96485 C/mol) Additional data: E° CuI/I-/Cu = –0.17 V E° Cd2+/Cd = –0.403 V
Chapter19: Applications Of Standard Electrode Potentials
Section: Chapter Questions
Problem 19.8QAP
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Question
Consider the
2CuI(s) + Cd(s) → Cd2+ + 2I– + 2Cu(s)
(Use R = 8.314 J/mol-K; F = 96485 C/mol)
Additional data:
E° CuI/I-/Cu = –0.17 V
E° Cd2+/Cd = –0.403 V
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