1) Using a table of standard electrode potentials (such as the one in your textbook), calculate the
standard cell potential for a cell made by placing a zinc electrode in a Zn2+(aq) solution which is
connected by an electrolyte to a Ag+(aq) solution containing a silver electrode. Show your work.
2) A student places a zinc electrode in a 0.80 M Zn2+(aq) solution which is connected by an
electrolyte to a 1.30 M Ag+(aq) solution containing a silver electrode. (Note that the solution
concentrations are not standard). Determine the initial voltage of the cell at 298 K. Show your
work. The Nernst equation will help you answer this question.
I don't see anything new here. See your other posts.
1) To calculate the standard cell potential, you need to use the standard electrode potentials (also known as standard reduction potentials) for each half-cell reaction and apply the equation:
Standard Cell Potential = E°(cathode) - E°(anode)
First, locate the standard reduction potentials for the half-reactions involved:
Zn2+(aq) + 2e- → Zn(s) with a standard reduction potential E°(Zn2+/Zn) = -0.76 V
Ag+(aq) + e- → Ag(s) with a standard reduction potential E°(Ag+/Ag) = 0.80 V
Next, substitute these values into the equation:
E°(cell) = E°(cathode) - E°(anode)
E°(cell) = 0.80 V - (-0.76 V)
E°(cell) = 1.56 V
Therefore, the standard cell potential for the cell is 1.56 V.
2) To determine the initial voltage of the cell using the Nernst equation, you need to consider the concentrations of the ions involved.
The Nernst equation is:
E = E° - (RT/nF) * ln(Q)
Where:
E = Cell potential
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of electrons transferred in the balanced equation
F = Faraday's constant (96,485 C/mol)
ln = Natural logarithm
Q = Reaction quotient (ratio of product concentrations to reactant concentrations)
In this case, the balanced half-reactions are unchanged, so n = 1.
The reaction quotient Q is determined by the concentrations of Zn2+ and Ag+ ions in solution.
Given:
[Zn2+] = 0.80 M
[Ag+] = 1.30 M
E°(cell) from part 1 = 1.56 V
Substituting the values into the Nernst equation:
E = E° - (RT/nF) * ln(Q)
E = 1.56 V - (8.314 J/(mol·K) * 298 K / (1 * 96,485 C/mol)) * ln((1.30 M)/(0.80 M))
Calculating the nested components:
ln((1.30 M)/(0.80 M)) = ln(1.625)
= 0.489
Substituting into the Nernst equation:
E = 1.56 V - (8.314 J/(mol·K) * 298 K / (1 * 96,485 C/mol)) * 0.489
Simplifying:
E = 1.56 V - 0.0254 V * 0.489
Calculating:
E ≈ 1.56 V - 0.0124 V
E ≈ 1.5476 V
Therefore, the initial voltage of the cell at 298 K is approximately 1.5476 V.