A voltaic cell is constructed that uses the following half-cell reactions.

Cu+(aq) + e- Cu(s)
I2(s) + 2 e- 2 I -(aq)
The cell is operated at 298 K with [Cu+ ] = 2.3 M and [I - ] = 3.3 M.

(a) Determine E for the cell at these concentrations.
(b) If [Cu+ ] were equal to 1.0 M, at what concentration of I - would the cell have zero potential?

What values are you using for Eo for I2 and Eo for Cu^+ ==> Cu

My values are so old they may not agree with yours from a more modern text.

To determine the standard cell potential (E°) for the voltaic cell, you need to find the difference in standard electrode potentials (E°) for the two half-cell reactions and add them together.

The standard electrode potential (E°) is a measure of the tendency of a half-cell reaction to occur at standard conditions. The values of E° are typically given in tables or can be calculated using the Nernst equation.

(a) To determine E° for the cell at the given concentrations, you need to calculate the cell potential using the Nernst equation:

E = E° - (RT / nF) * ln(Q),

where E is the cell potential, E° is the standard cell potential, R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin (298 K), n is the number of electrons transferred in the balanced equation (1 for both half-cell reactions), F is the Faraday constant (96485 C/mol), and ln(Q) is the natural logarithm of the reaction quotient.

The reaction quotient (Q) is calculated by dividing the product of the concentrations of the products (Cu+ for the first half-cell reaction and I- for the second half-cell reaction) by the product of the concentrations of the reactants (Cu for the first half-cell reaction and I2 for the second half-cell reaction).

For the first half-cell reaction:

Cu+(aq) + e- Cu(s)

The reaction quotient (Q1) is given by:

Q1 = [Cu+] / [Cu]

Substituting the given concentration values, you get:

Q1 = 2.3 M / 1 M = 2.3

For the second half-cell reaction:

I2(s) + 2 e- 2 I -(aq)

The reaction quotient (Q2) is given by:

Q2 = [I-]^2 / [I2]

Substituting the given concentration values, you get:

Q2 = (3.3 M)^2 / 1 = 10.89

Substituting the values into the Nernst equation:

E = E° - (RT / nF) * ln(Q)

E = E° - (8.314 J/mol*K * 298 K) / (1 * 96485 C/mol) * ln(2.3 / 10.89)

Now, you need to know the E° values for the half-cell reactions. Look them up in a table or get them from the question. Suppose the E° value for the first half-cell reaction is 0.34 V and for the second half-cell reaction is 0.54 V.

Substituting the E° values and solving for E:

E = 0.34 V - (8.314 J/mol*K * 298 K) / (1 * 96485 C/mol) * ln(2.3 / 10.89)

Calculate this to find the value of E.

(b) To find the concentration of I- at which the cell potential is zero, we set the cell potential (E) to zero in the Nernst equation and solve for Q, which represents the reaction quotient at that point. This will give us the concentration of I- required:

0 = E° - (RT / nF) * ln(Q)

0 = 0.34 V - (8.314 J/mol*K * 298 K) / (1 * 96485 C/mol) * ln(Q)

Solve this equation to find the value of Q, and then use Q to calculate the concentration of I-.