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The EMF of the following cell is 0.22 volt.
Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)
Which of the following will decrease the EMF of cell.
- a)Increasing pressure of H2(g) from 1 atm to 2 atm
- b)Increasing Cl_ concentration in Anodic compartment
- c)Increasing H+ concentration in cathodic compartment
- d)Decreasing KCl concentration in Anodic compartment.
Correct answer is option 'A,D'. Can you explain this answer?
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The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1...
Cell reaction
Anode
Ag(s) + Cl–1 → AgCl (s) + e–
Cathode 2H+ + 2e– → H2
Anode
Ag(s) + Cl–1 → AgCl (s) + e–
Cathode 2H+ + 2e– → H2
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The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1...
Explanation:
EMF of a cell is the difference in the electrode potentials of the cathode and the anode. Any change that affects the electrode potential of either cathode or anode will affect the EMF of the cell.
a) Increasing pressure of H2(g) from 1 atm to 2 atm
- The increase in pressure of H2(g) will shift the equilibrium of the reaction towards the side with fewer moles of gas, i.e., the reduction of H+ ions to H2 gas will be favored.
- This will cause an increase in the concentration of H+ ions in the cathodic compartment, which will increase the electrode potential of the cathode.
- As a result, the overall EMF of the cell will decrease.
d) Decreasing KCl concentration in Anodic compartment
- Decreasing the KCl concentration in the anodic compartment will decrease the concentration of Cl- ions available for the reaction at the anode.
- This will decrease the electrode potential of the anode.
- As a result, the overall EMF of the cell will decrease.
b) Increasing Cl- concentration in Anodic compartment
- Increasing the Cl- concentration in the anodic compartment will increase the concentration of Cl- ions available for the reaction at the anode.
- This will increase the electrode potential of the anode.
- As a result, the overall EMF of the cell will increase.
c) Increasing H+ concentration in cathodic compartment
- Increasing the H+ concentration in the cathodic compartment will increase the concentration of H+ ions available for the reduction reaction at the cathode.
- This will increase the electrode potential of the cathode.
- As a result, the overall EMF of the cell will increase.
Therefore, the correct options are A and D.
EMF of a cell is the difference in the electrode potentials of the cathode and the anode. Any change that affects the electrode potential of either cathode or anode will affect the EMF of the cell.
a) Increasing pressure of H2(g) from 1 atm to 2 atm
- The increase in pressure of H2(g) will shift the equilibrium of the reaction towards the side with fewer moles of gas, i.e., the reduction of H+ ions to H2 gas will be favored.
- This will cause an increase in the concentration of H+ ions in the cathodic compartment, which will increase the electrode potential of the cathode.
- As a result, the overall EMF of the cell will decrease.
d) Decreasing KCl concentration in Anodic compartment
- Decreasing the KCl concentration in the anodic compartment will decrease the concentration of Cl- ions available for the reaction at the anode.
- This will decrease the electrode potential of the anode.
- As a result, the overall EMF of the cell will decrease.
b) Increasing Cl- concentration in Anodic compartment
- Increasing the Cl- concentration in the anodic compartment will increase the concentration of Cl- ions available for the reaction at the anode.
- This will increase the electrode potential of the anode.
- As a result, the overall EMF of the cell will increase.
c) Increasing H+ concentration in cathodic compartment
- Increasing the H+ concentration in the cathodic compartment will increase the concentration of H+ ions available for the reduction reaction at the cathode.
- This will increase the electrode potential of the cathode.
- As a result, the overall EMF of the cell will increase.
Therefore, the correct options are A and D.
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The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer?
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The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer?.
The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The EMF of the following cell is 0.22 volt.Ag(s) |AgCl(s)|KCl(1M)|H+(1M)|H2(g) (1atm); Pt(s)Which of the following will decrease the EMF of cell.a)Increasing pressure of H2(g) from 1 atm to 2 atmb)Increasing Cl_concentration in Anodic compartmentc)Increasing H+concentration in cathodic compartmentd)Decreasing KCl concentration in Anodic compartment.Correct answer is option 'A,D'. Can you explain this answer?.
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