Question 8 Consider the electrochemical cell described by the following reaction at 310 K. The measured cell potential is 0.237 V. What is the concentration of iodide ion (in M) if there are 1.15 M of cadmium ions in the cell? 2Cul(s) + Cd(s) → Cd²+ + 21- +2Cu(s) (Use R = 8.314 J/mol-K; F = 96485 C/mol) Additional data: E° Cul/l-/Cu = -0.17 V E Cd2+/Cd = -0.403 V Express your answer with 2 decimal places and do not encode the units. If negative, include the sign in your encoded answer.

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Question 8
Consider the electrochemical cell described by the following reaction at 310 K. The measured cell potential is 0.237 V. What is the concentration of iodide ion (in M) if there are 1.15 M of cadmium ions in the cell?
2Cul(s) + Cd(s) → Cd²+ + 21¯ + 2Cu(s)
(Use R = 8.314 J/mol-K; F = 96485 C/mol)
Additional data:
E° Cul/l-/Cu = -0.17 V
E° Cd2+/Cd = -0.403 V
Express your answer with 2 decimal places and do not encode the units. If negative, include the sign in your encoded answer.
Transcribed Image Text:Question 8 Consider the electrochemical cell described by the following reaction at 310 K. The measured cell potential is 0.237 V. What is the concentration of iodide ion (in M) if there are 1.15 M of cadmium ions in the cell? 2Cul(s) + Cd(s) → Cd²+ + 21¯ + 2Cu(s) (Use R = 8.314 J/mol-K; F = 96485 C/mol) Additional data: E° Cul/l-/Cu = -0.17 V E° Cd2+/Cd = -0.403 V Express your answer with 2 decimal places and do not encode the units. If negative, include the sign in your encoded answer.
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