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pH of the solution in the anode compartment of the follwoing cell at 298 K is x
when Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.
Derive the value of x.
Correct answer is '1'. Can you explain this answer?
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pH of the solution in the anode compartment of the follwoing cell at 2...
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pH of the solution in the anode compartment of the follwoing cell at 2...
Given Information:
- Ecell - Ecell = 0.0591 V
- Pt (H2) | pH = x|| Ni2 (1M)| Ni
To Find:
The value of pH in the anode compartment.
Explanation:
The given cell consists of two half-cells: the anode compartment and the cathode compartment.
Half-Reactions:
At the anode (oxidation):
Ni(s) → Ni2+(aq) + 2e-
At the cathode (reduction):
2H+(aq) + 2e- → H2(g)
The overall reaction is the sum of the half-reactions:
Ni(s) + 2H+(aq) → Ni2+(aq) + H2(g)
The standard potential of this cell can be calculated using the Nernst equation:
Ecell = E°cell - (0.0591 V / n) * log(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of moles of electrons transferred
Q is the reaction quotient
In this case, since the cell is at equilibrium (Ecell - Ecell = 0), the reaction quotient is equal to 1.
Deriving the Value of x:
Let's assume the pH in the anode compartment is x.
Since Pt (H2) is used as the electrode in the cathode compartment, the concentration of H+ ions is 1 M in both compartments. Therefore, the reaction quotient Q can be calculated as follows:
Q = [Ni2+(aq)] / [H+(aq)]^2
At equilibrium, Q = 1. Thus, we have:
1 = [Ni2+(aq)] / [H+(aq)]^2
Since the concentration of H+ ions is 1 M, we can simplify the equation to:
1 = [Ni2+(aq)]
Therefore, the concentration of Ni2+ ions in the anode compartment is 1 M.
Since the pH is equal to the negative logarithm of the hydrogen ion concentration (pH = -log[H+]), the value of x is 1.
Conclusion:
The value of pH in the anode compartment is 1.
- Ecell - Ecell = 0.0591 V
- Pt (H2) | pH = x|| Ni2 (1M)| Ni
To Find:
The value of pH in the anode compartment.
Explanation:
The given cell consists of two half-cells: the anode compartment and the cathode compartment.
Half-Reactions:
At the anode (oxidation):
Ni(s) → Ni2+(aq) + 2e-
At the cathode (reduction):
2H+(aq) + 2e- → H2(g)
The overall reaction is the sum of the half-reactions:
Ni(s) + 2H+(aq) → Ni2+(aq) + H2(g)
The standard potential of this cell can be calculated using the Nernst equation:
Ecell = E°cell - (0.0591 V / n) * log(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of moles of electrons transferred
Q is the reaction quotient
In this case, since the cell is at equilibrium (Ecell - Ecell = 0), the reaction quotient is equal to 1.
Deriving the Value of x:
Let's assume the pH in the anode compartment is x.
Since Pt (H2) is used as the electrode in the cathode compartment, the concentration of H+ ions is 1 M in both compartments. Therefore, the reaction quotient Q can be calculated as follows:
Q = [Ni2+(aq)] / [H+(aq)]^2
At equilibrium, Q = 1. Thus, we have:
1 = [Ni2+(aq)] / [H+(aq)]^2
Since the concentration of H+ ions is 1 M, we can simplify the equation to:
1 = [Ni2+(aq)]
Therefore, the concentration of Ni2+ ions in the anode compartment is 1 M.
Since the pH is equal to the negative logarithm of the hydrogen ion concentration (pH = -log[H+]), the value of x is 1.
Conclusion:
The value of pH in the anode compartment is 1.
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pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer?.
pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer?.
Solutions for pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer?, a detailed solution for pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? has been provided alongside types of pH of the solution in the anode compartment of the follwoing cell at 298 K is xwhen Ecell - E°cell = 0.0591 V , Pt (H2) | pH = x|| Ni2+ (1M)| Ni.Derive the value of x. Correct answer is '1'. Can you explain this answer? theory, EduRev gives you an
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