Class 12 Exam > Class 12 Questions > For the cell reaction, [1998] Cu2+ (C1, aq) +...
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For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function of
- a)ln (C1)
- b)ln (C2/C1)
- c)ln (C2)
- d)ln (C1 + C2)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + ...
For concentration cell,
In it R, T, n and F are constant So E is based upon ln C2 / C1
Now 
= –RTlnC2/C1
At constant temperature ΔG is based upon ln C2/C1.
Most Upvoted Answer
For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + ...
ΔG, can be calculated using the equation:
ΔG = -nFE
where n is the number of moles of electrons transferred in the balanced chemical equation, F is Faraday's constant (96,485 C/mol), and E is the cell potential.
To calculate ΔG, we need to determine the number of moles of electrons transferred. Looking at the balanced equation, we see that 2 moles of electrons are transferred for every 1 mole of Cu2+ and Zn2+ ions.
Next, we need to find the cell potential (E). This can be determined using the standard reduction potentials for the half-reactions involved. The standard reduction potentials can be found in a reference table.
The half-reactions for this cell are:
Cu2+(aq) + 2e- = Cu(s) (reduction half-reaction)
Zn(s) = Zn2+(aq) + 2e- (oxidation half-reaction)
The standard reduction potential for the Cu2+(aq) + 2e- = Cu(s) half-reaction is +0.34 V.
The standard reduction potential for the Zn(s) = Zn2+(aq) + 2e- half-reaction is -0.76 V.
To calculate the cell potential (E), we subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction:
E = E(reduction) - E(oxidation)
E = 0.34 V - (-0.76 V)
E = 1.10 V
Now we have all the necessary values to calculate ΔG:
ΔG = -nFE
ΔG = -(2 mol)(96,485 C/mol)(1.10 V)
ΔG = -212,267 C
Since ΔG is in units of coulombs (C), we can convert it to joules (J) by using the conversion factor 1 C = 1 J:
ΔG = -212,267 J
Therefore, the change in free energy for the given cell reaction is -212,267 J.
ΔG = -nFE
where n is the number of moles of electrons transferred in the balanced chemical equation, F is Faraday's constant (96,485 C/mol), and E is the cell potential.
To calculate ΔG, we need to determine the number of moles of electrons transferred. Looking at the balanced equation, we see that 2 moles of electrons are transferred for every 1 mole of Cu2+ and Zn2+ ions.
Next, we need to find the cell potential (E). This can be determined using the standard reduction potentials for the half-reactions involved. The standard reduction potentials can be found in a reference table.
The half-reactions for this cell are:
Cu2+(aq) + 2e- = Cu(s) (reduction half-reaction)
Zn(s) = Zn2+(aq) + 2e- (oxidation half-reaction)
The standard reduction potential for the Cu2+(aq) + 2e- = Cu(s) half-reaction is +0.34 V.
The standard reduction potential for the Zn(s) = Zn2+(aq) + 2e- half-reaction is -0.76 V.
To calculate the cell potential (E), we subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction:
E = E(reduction) - E(oxidation)
E = 0.34 V - (-0.76 V)
E = 1.10 V
Now we have all the necessary values to calculate ΔG:
ΔG = -nFE
ΔG = -(2 mol)(96,485 C/mol)(1.10 V)
ΔG = -212,267 C
Since ΔG is in units of coulombs (C), we can convert it to joules (J) by using the conversion factor 1 C = 1 J:
ΔG = -212,267 J
Therefore, the change in free energy for the given cell reaction is -212,267 J.
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For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function ofa)ln (C1)b)ln (C2/C1)c)ln (C2)d)ln (C1 + C2)Correct answer is option 'B'. 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 For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function ofa)ln (C1)b)ln (C2/C1)c)ln (C2)d)ln (C1 + C2)Correct answer is option 'B'. 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 For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function ofa)ln (C1)b)ln (C2/C1)c)ln (C2)d)ln (C1 + C2)Correct answer is option 'B'. Can you explain this answer?.
For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function ofa)ln (C1)b)ln (C2/C1)c)ln (C2)d)ln (C1 + C2)Correct answer is option 'B'. 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 For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function ofa)ln (C1)b)ln (C2/C1)c)ln (C2)d)ln (C1 + C2)Correct answer is option 'B'. 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 For the cell reaction, [1998] Cu2+ (C1, aq) + Zn(s) = Zn2+ (C2, aq) + Cu(s) of an electrochemical cell, the change in free energy, ΔG, at a given temperature is a function ofa)ln (C1)b)ln (C2/C1)c)ln (C2)d)ln (C1 + C2)Correct answer is option 'B'. Can you explain this answer?.
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