Calculate the E.M.F. of the following cell at 298 K : Mg Mg 2 + 0 . 130 M Ag + 1 . 0 × 10 - 4 M Ag Given that E Mg 2 + ∣ Mg Θ = - 2 . 37 V , E Ag + ∣ Ag Θ = 0 . 80 V

Given, E Mg 2 + ∣ Mg ° = - 2 . 37 V ( anode ) , E Ag + ∣ Ag ° = 0 . 80 V ( Cathode ) Applying Nernst equation, E cell = E cell ° - 0 . 0591 V 2 log Mg 2 + Ag + 2 E cell = E cell ° - 0 . 0591 2 log 0 . 130 M 1 . 0 × 10 - 4 M 2 where , E cell = E cathode ° - E anode ° E cell = 0 . 80 - - 2 . 37 E cell = 3 . 17 V E cell = 3 . 17 - 0 . 0591 2 log 0 . 130 M 1 . 0 × 10 - 4 M 2 E cell = 2 . 96 V

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NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 3 - Electrochemistry>Follow Up Problems

K L Chugh Solutions for Chapter: Electrochemistry, Exercise 7: Follow Up Problems

Author:K L Chugh

K L Chugh Chemistry Solutions for Exercise - K L Chugh Solutions for Chapter: Electrochemistry, Exercise 7: Follow Up Problems

Attempt the practice questions on Chapter 3: Electrochemistry, Exercise 7: Follow Up Problems with hints and solutions to strengthen your understanding. NEW SYSTEMATIC MODERN CHEMISTRY solutions are prepared by Experienced Embibe Experts.