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A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)?
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A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn...
Calculation of EMF of the Cell:
To calculate the EMF of the cell, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell = EMF of the cell
E°cell = Standard EMF of the cell
n = Number of electrons transferred in the balanced equation
Q = Reaction quotient
Chemical Equation for the Electrode Reactions:
The electrode reactions can be represented as follows:
At the anode (oxidation half-reaction):
Zn(s) → Zn2+(aq) + 2e-
At the cathode (reduction half-reaction):
Pb2+(aq) + 2e- → Pb(s)
Overall cell reaction:
Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)
Representation of the Cell:
The cell can be represented as follows:
Anode: Zn(s) | Zn2+(aq) || Pb2+(aq) | Pb(s) : Cathode
Calculating the Reaction Quotient (Q):
To calculate the reaction quotient, we need to determine the concentrations of the species involved in the cell reaction.
Given:
0.1 M Zn(NO3)2 solution
0.02 M Pb(NO3)2 solution
Since the concentrations of the metal ions are the same as the concentration of their respective nitrate salts, the reaction quotient can be calculated as follows:
Q = [Zn2+]/[Pb2+]
Q = (0.1)/(0.02)
Q = 5
Calculating the EMF of the Cell:
Given:
E°Zn2+/Zn = -0.76 V
E°Pb2+/Pb = -0.13 V
Using the Nernst equation, we can calculate the EMF of the cell:
Ecell = E°cell - (0.0592/n) * log(Q)
The number of electrons transferred in the balanced equation is 2, so:
n = 2
Ecell = -0.76 V - (0.0592/2) * log(5)
Ecell = -0.76 V - (0.0296) * log(5)
Ecell = -0.76 V - 0.0296 * 0.69
Ecell = -0.76 V - 0.0204
Ecell = -0.7804 V
Therefore, the EMF of the cell is -0.7804 V.
To calculate the EMF of the cell, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell = EMF of the cell
E°cell = Standard EMF of the cell
n = Number of electrons transferred in the balanced equation
Q = Reaction quotient
Chemical Equation for the Electrode Reactions:
The electrode reactions can be represented as follows:
At the anode (oxidation half-reaction):
Zn(s) → Zn2+(aq) + 2e-
At the cathode (reduction half-reaction):
Pb2+(aq) + 2e- → Pb(s)
Overall cell reaction:
Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)
Representation of the Cell:
The cell can be represented as follows:
Anode: Zn(s) | Zn2+(aq) || Pb2+(aq) | Pb(s) : Cathode
Calculating the Reaction Quotient (Q):
To calculate the reaction quotient, we need to determine the concentrations of the species involved in the cell reaction.
Given:
0.1 M Zn(NO3)2 solution
0.02 M Pb(NO3)2 solution
Since the concentrations of the metal ions are the same as the concentration of their respective nitrate salts, the reaction quotient can be calculated as follows:
Q = [Zn2+]/[Pb2+]
Q = (0.1)/(0.02)
Q = 5
Calculating the EMF of the Cell:
Given:
E°Zn2+/Zn = -0.76 V
E°Pb2+/Pb = -0.13 V
Using the Nernst equation, we can calculate the EMF of the cell:
Ecell = E°cell - (0.0592/n) * log(Q)
The number of electrons transferred in the balanced equation is 2, so:
n = 2
Ecell = -0.76 V - (0.0592/2) * log(5)
Ecell = -0.76 V - (0.0296) * log(5)
Ecell = -0.76 V - 0.0296 * 0.69
Ecell = -0.76 V - 0.0204
Ecell = -0.7804 V
Therefore, the EMF of the cell is -0.7804 V.
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A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)?
Question Description
A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)?.
A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E�Zn2+/Zn = -0.76 V; E�Pb2+/Pb = -0.13 V)?.
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