For the reduction of Cu2+by Zn, ∆G◦=−212 kJ/mol and E◦= +1.10 V. If the
coefficients in the chemical equation for this reaction are multiplied by 2, ∆G◦=−424 kJ/mol. This means E◦= +2.20 V.
True or false?
false.
E for the rxn is 1.10v and that is true with a little Zn and a little Cu^2+ or a big log of each.
false
False. The relationship between ∆G◦ and E◦ is given by the equation ∆G◦ = -nFE◦, where n is the number of electrons transferred in the reaction and F is the Faraday constant. Since the coefficients in the chemical equation have been multiplied by 2, it means that n has doubled. Therefore, the new ∆G◦ would be -2 times the original ∆G◦, which is -2 * (-212 kJ/mol) = +424 kJ/mol. This means that E◦ would be -2 * 1.10 V = -2.20 V. The statement that E◦ = +2.20 V is incorrect.
True.
To explain:
The standard Gibbs free energy change (∆G◦) is related to the standard cell potential (E◦) using the equation:
∆G◦ = -nF E◦
where:
- ∆G◦ is the standard Gibbs free energy change
- n is the number of electrons transferred in the balanced equation
- F is Faraday's constant (96,485 C/mol)
- E◦ is the standard cell potential
In the given equation, Cu2+ is being reduced to Cu by Zn. The balanced equation is:
Cu2+ + Zn -> Cu + Zn2+
Given that ∆G◦ = -212 kJ/mol and E◦ = +1.10 V, let's calculate the number of electrons transferred (n) using the equation:
∆G◦ = -nF E◦
-212,000 J/mol = -(n)(96,485 C/mol)(1.10 V)
Solving for n, we find n = 2.
Now, if we multiply the coefficients in the balanced equation by 2, the new balanced equation becomes:
2Cu2+ + 2Zn -> 2Cu + 2Zn2+
In this new equation, the number of electrons transferred is 2n = 4.
Using the same equation as before, with the new values for ∆G◦ (-424 kJ/mol) and n (4), we can calculate the new E◦:
-424,000 J/mol = -(4)(96,485 C/mol)(E◦)
Solving for E◦, we find E◦ ≈ 2.20 V.
Therefore, it is true that when the coefficients in the chemical equation for the reaction are multiplied by 2, ∆G◦ = -424 kJ/mol, meaning E◦ = +2.20 V.