Calculate the \of the galvanic cell containing the copper electrode and the zinc electrode where one is immersed in1.5 M CuSO4 solution and the other in 0.01 M ZnSO4 solution (E(Zn=-0.76 V, E(Cu=+0.34 V)
Zn ==> Zn^2+ + 2e .........Eo ox = +0.76
Cu^2+ + 2e ==> Cu..........Eo red = +0.34
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Zn(s) + Cu^2+(1M) ==> Zn^2+(1M) + Cu(s) .......Eo cell (std) = 1.10 v
Ecell = Eocell std - (0.059/n)*log[(Zn^2)(Cu)/(Zn)(Cu^2+)]
So substitute and solve for Ecell.
Eo cell std is the 1.10 v
n = total electrons exchanged = 2
(Zn) = 1.0 and (Cu) = 1.0 by definition
(Zn^2+) = 0.01 M from the problem.
(Cu^2+) = 1.5 M from the problem.
Post your work if you get stuck.
To calculate the potential (Ecell) of the galvanic cell, we can use the formula:
Ecell = E(reduction) - E(oxidation)
First, we need to determine which electrode is undergoing reduction and which is undergoing oxidation.
Given that E(Zn) = -0.76 V and E(Cu) = +0.34 V, we can compare the values to determine the reduction and oxidation reactions:
The more positive standard reduction potential (E) indicates the electrode undergoing reduction, while the more negative E indicates the electrode undergoing oxidation.
In this case, Cu has a more positive E value (+0.34 V), so it will undergo reduction, while Zn (-0.76 V) will undergo oxidation.
Therefore, the half-reactions can be written as follows:
Reduction: Cu2+ + 2e- → Cu (+0.34 V)
Oxidation: Zn → Zn2+ + 2e- (-0.76 V)
Now, we can calculate the overall Ecell by subtracting the oxidation potential from the reduction potential:
Ecell = E(reduction) - E(oxidation)
Ecell = (+0.34 V) - (-0.76 V)
Ecell = +1.10 V
So, the potential of the galvanic cell containing the copper electrode and the zinc electrode is +1.10 V.
To calculate the cell potential of a galvanic cell, you need to use the Nernst equation, which relates the cell potential to the concentrations of the species involved.
The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the gas constant (8.314 J/(mol K))
T is the temperature in Kelvin
n is the number of electrons transferred in the balanced cell reaction
F is Faraday's constant (96485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
Let's start calculating the standard cell potential (E°cell). It is given by the difference between the standard reduction potentials for the two half-reactions involved in the cell:
E°cell = E°cathode - E°anode
Given:
E°(Zn) = -0.76 V (reduction potential for zinc)
E°(Cu) = +0.34 V (reduction potential for copper)
E°cell = E°cathode - E°anode
E°cell = E°(Cu) - E°(Zn)
E°cell = +0.34 V - (-0.76 V)
E°cell = +0.34 V + 0.76 V
E°cell = +1.10 V
Now, let's calculate the cell potential (Ecell) using the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
The cell reaction can be written as:
Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq)
The reaction quotient (Q) is the ratio of the concentrations of the products to the reactants, raised to the power of their stoichiometric coefficients. In this case, it simplifies to the ratio of the concentration of Zn2+ to the concentration of Cu2+:
Q = [Zn2+]/[Cu2+]
Given:
[Zn2+] = 0.01 M (concentration of Zn2+)
[Cu2+] = 1.5 M (concentration of Cu2+)
Q = [Zn2+]/[Cu2+]
Q = 0.01 M / 1.5 M
Q = 0.0067
Now, let's substitute the values into the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Ecell = 1.10 V - (8.314 J/(mol K) * T) / (2 * 96485 C/mol) * ln(0.0067)
You will need to provide information about the temperature (T) in Kelvin in order to calculate Ecell.