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The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V is
- a)+ 0.44 V
- b)+ 1.12 V
- c)- 0.44 V
- d)- 1.12 V
Correct answer is option 'D'. Can you explain this answer?
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The e.m.f. of a galvanic cell with electrode potentials of Al equal to...
Galvanic Cell:
A galvanic cell consists of two half-cells, each of which contains an electrode immersed in an electrolyte solution. The two half-cells are connected through a wire and a salt bridge.
Electrode Potential:
The electrode potential is the measure of the tendency of a metal electrode to lose or gain electrons. The standard electrode potential is a measure of the tendency of a metal electrode to lose electrons when it is in contact with its own aqueous ions.
Calculation of e.m.f.:
The e.m.f. of a galvanic cell is given by the difference in the electrode potentials of the two half-cells. The half-cell with the higher electrode potential will act as the cathode, and the half-cell with the lower electrode potential will act as the anode.
Given:
Electrode potential of Al = -1.66 V
Electrode potential of Mg = -0.54 V
The half-cell with the higher electrode potential is Mg, and it will act as the cathode.
The half-cell with the lower electrode potential is Al, and it will act as the anode.
Therefore, the e.m.f. of the galvanic cell will be:
e.m.f. = (Electrode potential of cathode) - (Electrode potential of anode)
e.m.f. = (-0.54 V) - (-1.66 V)
e.m.f. = 1.12 V
Hence, the correct option is (D) -1.12 V.
A galvanic cell consists of two half-cells, each of which contains an electrode immersed in an electrolyte solution. The two half-cells are connected through a wire and a salt bridge.
Electrode Potential:
The electrode potential is the measure of the tendency of a metal electrode to lose or gain electrons. The standard electrode potential is a measure of the tendency of a metal electrode to lose electrons when it is in contact with its own aqueous ions.
Calculation of e.m.f.:
The e.m.f. of a galvanic cell is given by the difference in the electrode potentials of the two half-cells. The half-cell with the higher electrode potential will act as the cathode, and the half-cell with the lower electrode potential will act as the anode.
Given:
Electrode potential of Al = -1.66 V
Electrode potential of Mg = -0.54 V
The half-cell with the higher electrode potential is Mg, and it will act as the cathode.
The half-cell with the lower electrode potential is Al, and it will act as the anode.
Therefore, the e.m.f. of the galvanic cell will be:
e.m.f. = (Electrode potential of cathode) - (Electrode potential of anode)
e.m.f. = (-0.54 V) - (-1.66 V)
e.m.f. = 1.12 V
Hence, the correct option is (D) -1.12 V.
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The e.m.f. of a galvanic cell with electrode potentials of Al equal to...
Calculation of e.m.f of the galvanic cell
Given electrode potentials:
• E°(Al) = -1.66 V
• E°(Mg) = -0.54 V
The e.m.f of the galvanic cell can be calculated using the Nernst equation
Ecell = E°cell - (RT/nF) lnQ
where,
• E°cell is the standard electrode potential of the cell.
• R is the gas constant (8.314 J K-1 mol-1).
• T is the temperature in Kelvin.
• n is the number of electrons transferred in the cell reaction.
• F is the Faraday constant (96,485 C mol-1).
• Q is the reaction quotient.
The reaction taking place in the cell is:
Al(s) + Mg2+(aq) → Al3+(aq) + Mg(s)
The cell reaction can be split up into two half-cells:
Al3+(aq) + 3e- → Al(s) (reduction half-cell)
Mg2+(aq) + 2e- → Mg(s) (oxidation half-cell)
The standard cell potential can be calculated using:
E°cell = E°reduction - E°oxidation
E°cell = (-1.66 V) - (-0.54 V)
E°cell = -1.12 V
The reaction quotient Q can be calculated using the concentrations of the reactants and products:
Q = [Al3+]/[Mg2+]
At equilibrium, Q = Kc, where Kc is the equilibrium constant.
Assuming the concentrations of Al3+ and Mg2+ are both 1 M, the value of Q can be calculated as:
Q = [Al3+]/[Mg2+] = 1/1 = 1
Using the values of E°cell, R, T, n, F, and Q, the e.m.f of the galvanic cell can be calculated as:
Ecell = E°cell - (RT/nF) lnQ
Ecell = -1.12 V - [(8.314 J K-1 mol-1) (298 K) / (3 mol e-)(96,485 C mol-1)] ln(1)
Ecell = -1.12 V
Therefore, the e.m.f of the galvanic cell is -1.12 V.
The correct answer is option (d) -1.12 V
Given electrode potentials:
• E°(Al) = -1.66 V
• E°(Mg) = -0.54 V
The e.m.f of the galvanic cell can be calculated using the Nernst equation
Ecell = E°cell - (RT/nF) lnQ
where,
• E°cell is the standard electrode potential of the cell.
• R is the gas constant (8.314 J K-1 mol-1).
• T is the temperature in Kelvin.
• n is the number of electrons transferred in the cell reaction.
• F is the Faraday constant (96,485 C mol-1).
• Q is the reaction quotient.
The reaction taking place in the cell is:
Al(s) + Mg2+(aq) → Al3+(aq) + Mg(s)
The cell reaction can be split up into two half-cells:
Al3+(aq) + 3e- → Al(s) (reduction half-cell)
Mg2+(aq) + 2e- → Mg(s) (oxidation half-cell)
The standard cell potential can be calculated using:
E°cell = E°reduction - E°oxidation
E°cell = (-1.66 V) - (-0.54 V)
E°cell = -1.12 V
The reaction quotient Q can be calculated using the concentrations of the reactants and products:
Q = [Al3+]/[Mg2+]
At equilibrium, Q = Kc, where Kc is the equilibrium constant.
Assuming the concentrations of Al3+ and Mg2+ are both 1 M, the value of Q can be calculated as:
Q = [Al3+]/[Mg2+] = 1/1 = 1
Using the values of E°cell, R, T, n, F, and Q, the e.m.f of the galvanic cell can be calculated as:
Ecell = E°cell - (RT/nF) lnQ
Ecell = -1.12 V - [(8.314 J K-1 mol-1) (298 K) / (3 mol e-)(96,485 C mol-1)] ln(1)
Ecell = -1.12 V
Therefore, the e.m.f of the galvanic cell is -1.12 V.
The correct answer is option (d) -1.12 V
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The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer?
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The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer?.
The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The e.m.f. of a galvanic cell with electrode potentials of Al equal to -1.66 V and that of Mg equal to - 0.54 V isa)+ 0.44 Vb)+ 1.12 Vc)- 0.44 Vd)- 1.12 VCorrect answer is option 'D'. Can you explain this answer?.
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