Published by:
CGP EDU Academic Team
Published on: August 12, 2026
Consider the following electrochemical cell :
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
The spontaneous cell reaction : Zn + 2Ag + (aq)
Zn 2+ (aq) + 2Ag (s)
1.56 V [Zn 2+ ] = 4 × 10 –4 M
As we add KI to cathode chamber, some Ag + will precipitate out as :
Ag + + I – ⎯→ AgI
The above reaction reducing [Ag + ] from cathode chamber. This will reduce E cell according to Nernst’s equation.
Sol. Zn + 2Ag + (aq)
Zn 2+ (aq) + 2Ag(s)
Eº = 0.8 – (– 0.76) = 1.56 V
1.6 = 1.56–
log
or ( ;k ) 1.356 = log 
or ( ;k ) log
= 1.356 or ( ;k )
= 22.7 or ( ;k ) [Zn 2+ ] =
= 4.4 × 10 –4 M
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
In the cell the negative terminal is
Which of the following reaction is possible at anode?
For a certain redox reaction, is positive this means that
How much electricity in terms of Faraday is required to produce 100 g of Ca from molten
In electrolysis of dilute what is liberated at anode?
The quantity of charge required to obtain one mole of aluminum from is _______.