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Consider the following relations for emf of a electrochemical cell: [2010]
(i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)
Which of the above relations are correct?
(i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)
Which of the above relations are correct?
- a)(ii) and (iv)
- b)(iii) and (i)
- c)(i) and (ii)
- d)(iii) and (iv)
Correct answer is option 'A'. Can you explain this answer?
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Consider the following relations for emf of a electrochemical cell: ...
Option (b) and (d) are correct
∴ Correct choice : (a)
∴ Correct choice : (a)
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Consider the following relations for emf of a electrochemical cell: ...
- (Reduction potential of cathode)
This relation is known as the Gibbs-Helmholtz equation for the emf of an electrochemical cell. It is based on the thermodynamic principle that the energy released during the oxidation reaction at the anode is equal to the energy required for the reduction reaction at the cathode. The difference in the two potentials determines the net emf of the cell.
(ii) emf of cell = E°(cathode) - E°(anode)
This relation is based on the standard reduction potentials of the half reactions at the cathode and anode. E° is the standard electrode potential, which is the potential difference between an electrode and a reference electrode under standard conditions. The emf of the cell is determined by the difference in the standard electrode potentials of the cathode and anode.
Both these relations are used to calculate the emf of a electrochemical cell and provide useful information about the energy changes that occur during the electrochemical reaction.
This relation is known as the Gibbs-Helmholtz equation for the emf of an electrochemical cell. It is based on the thermodynamic principle that the energy released during the oxidation reaction at the anode is equal to the energy required for the reduction reaction at the cathode. The difference in the two potentials determines the net emf of the cell.
(ii) emf of cell = E°(cathode) - E°(anode)
This relation is based on the standard reduction potentials of the half reactions at the cathode and anode. E° is the standard electrode potential, which is the potential difference between an electrode and a reference electrode under standard conditions. The emf of the cell is determined by the difference in the standard electrode potentials of the cathode and anode.
Both these relations are used to calculate the emf of a electrochemical cell and provide useful information about the energy changes that occur during the electrochemical reaction.
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Consider the following relations for emf of a electrochemical cell: ...
Electro-chemical series -
The standard reduction potential of a large number of electrodes have been measured using standard hydrogen electrode as the reference electrode. These various electrode can be arranged in increasing electrode potential.
EMF of a cell = Reduction potential of cathode- reduction potential of anode
= Reduction potential of cathode +oxidation potential of anode
= Oxidation potential of anode -oxidation potential of cathode.
Option 1)
2 and 4
This is correct option
The standard reduction potential of a large number of electrodes have been measured using standard hydrogen electrode as the reference electrode. These various electrode can be arranged in increasing electrode potential.
EMF of a cell = Reduction potential of cathode- reduction potential of anode
= Reduction potential of cathode +oxidation potential of anode
= Oxidation potential of anode -oxidation potential of cathode.
Option 1)
2 and 4
This is correct option
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Consider the following relations for emf of a electrochemical cell: [2010](i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)Which of the above relations are correct?a)(ii) and (iv)b)(iii) and (i)c)(i) and (ii)d)(iii) and (iv)Correct answer is option 'A'. Can you explain this answer?
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Consider the following relations for emf of a electrochemical cell: [2010](i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)Which of the above relations are correct?a)(ii) and (iv)b)(iii) and (i)c)(i) and (ii)d)(iii) and (iv)Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Consider the following relations for emf of a electrochemical cell: [2010](i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)Which of the above relations are correct?a)(ii) and (iv)b)(iii) and (i)c)(i) and (ii)d)(iii) and (iv)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the following relations for emf of a electrochemical cell: [2010](i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)Which of the above relations are correct?a)(ii) and (iv)b)(iii) and (i)c)(i) and (ii)d)(iii) and (iv)Correct answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice Consider the following relations for emf of a electrochemical cell: [2010](i) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)(ii) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)(iii) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)(iv) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)Which of the above relations are correct?a)(ii) and (iv)b)(iii) and (i)c)(i) and (ii)d)(iii) and (iv)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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