NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
A 1.0 M solution of Cd 2+ is added to excess iron and the system is allowed to reach equilibrium. What is the concentration in mol of Cd 2+ ? Cd 2 + aq + Fe s → Cd s + Fe 2 + aq ; E o = 0.037 V Given: log 18 = 1.25 Report your answer upto two decimal places.
Correct answer
0.05
Step-by-step solution
Cd 2 + aq + Fe s ⇌ Cd s + Fe 2 + aq At eqm. 1 - x - - x At equilibrium, E o = 0.0591 2 log Fe 2 + Cd 2 + 0.037 = 0.0591 2 log x 1 - x x = Fe 2 + ⇒ 0.947 M ∴ Cd 2 + = 0.053 M