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Class 12 Exam  >  Class 12 Questions  >  Following cell has EMF 0.7995 V.Pt | H2 (1 at... Start Learning for Free
Following cell has EMF 0.7995 V.
Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (1M) | Ag
If we add enough KCl to the Ag cell so that the final Cl- is 1M. Now the measured emf of the cell is 0.222 V.
The Ksp of AgCl would be :
  • a)
    1 x 10 – 9.8 
  • b)
    1x10 – 19.6 
  • c)
    2x10– 10   
  • d)
    2.64x10–14
Correct answer is option 'A'. Can you explain this answer?
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Following cell has EMF 0.7995 V.Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (...
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Following cell has EMF 0.7995 V.Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (...
Given:
- EMF of the cell = 0.222 V
- Initial EMF of the cell = 0.7995 V

To Determine:
The Ksp of AgCl

Explanation:
The given cell is a concentration cell, where the concentration of Ag+ on one side is higher than the other side. The measured EMF of the cell is lower than the initial EMF, indicating a decrease in the concentration of Ag+ ions.

Step 1: Identifying the Reduction Half-Reaction:
To understand the reduction half-reaction, let's write the overall cell reaction:

Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (1M) | Ag

The reduction half-reaction in this cell is the reduction of Ag+ ions to Ag:

Ag+ + e- → Ag

Step 2: Determining the Change in Concentration:
Since the measured EMF is lower than the initial EMF, it suggests a decrease in the concentration of Ag+ ions. This decrease in concentration is due to the addition of KCl, which forms AgCl precipitate by reacting with Ag+ ions.

The addition of KCl increases the concentration of Cl- ions. The Ag+ ions react with the Cl- ions to form AgCl precipitate according to the reaction:

Ag+ + Cl- → AgCl

Therefore, the change in concentration of Ag+ ions is equal to the concentration of Cl- ions added, which is 1 M.

Step 3: Applying the Nernst Equation:
The Nernst equation relates the measured EMF of the cell (Ecell) to the standard EMF of the cell (E°cell) and the reaction quotient (Q):

Ecell = E°cell - (0.0592/n)log(Q)

Since the cell is at equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (Ksp) for the precipitation reaction:

Q = Ksp

Therefore, we can rewrite the Nernst equation as:

Ecell = E°cell - (0.0592/n)log(Ksp)

Step 4: Solving for Ksp:
We can rearrange the Nernst equation to solve for Ksp:

Ksp = 10^((E°cell - Ecell) * (n/0.0592))

Substituting the given values:

E°cell = 0.7995 V
Ecell = 0.222 V
n = 1 (since one electron is transferred in the reduction half-reaction)

Ksp = 10^((0.7995 - 0.222) * (1/0.0592))
= 10^(0.577 / 0.0592)
≈ 10^(9.75)
≈ 1 × 10^9.75
≈ 1 × 10^9.8

Therefore, the Ksp of AgCl is approximately 1 × 10^9.8, which corresponds to option A.
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Following cell has EMF 0.7995 V.Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (1M) | AgIf we add enough KCl to the Ag cell so that the final Cl- is 1M. Now the measured emf of the cell is 0.222 V.The Ksp of AgCl would be :a)1 x10 – 9.8b)1x10 – 19.6c)2x10– 10d)2.64x10–14Correct answer is option 'A'. Can you explain this answer?
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