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+ ] = 2.7 M and [I− ] = 2.7 M.

(a) Determine E for the cell at these concentrations.

(b) If [Cu+ ] was equal to 1.2 M, at what concentration of I− would the cell have zero potential?

When you post problems that depend upon numbers you would do well to post the numbers in your text because texts change over the years and those constants change. The E value I looked up for I is =0.535 and the E value for Cu is +0.521.

Step 1 is to use the reduction ernst equation, substitute 2.7M and calculate E for I. I get +.509 but your values may be different. Do the same for Cu and I obtained 0.469; again your numbers may be different.
Then write the equation.
I2 ==> 2I^- E = +0.509v
Cu + e ==> Cu E = -0.469v
----------------------
I2 + 2Cu ==> 2Cu^+ + 2I^- Ecell = sum of E values.

To do b I would use the overall equation.
Ecell = EoCell - (0.0592/2)logQ
log Q = (I^-)^2(Cu^+)^2/((Cu)(I2) and solve for I^-

To determine the cell potential, we can use the Nernst equation:

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 half-reaction
F = Faraday's constant (96,485 C/mol)
Q = reaction quotient

First, let's calculate the cell potential E for the given concentrations.

(a) Determine E for the cell at these concentrations:
The balanced cell reaction is:
Cu+(aq) + I2(s) -> Cu(s) + 2 I−(aq)

Looking at the half-cell reactions:
Cu+(aq) + e− -> Cu(s) (E° = 0.52 V)
I2(s) + 2 e− -> 2 I−(aq) (E° = 0.54 V)

Since the reaction is spontaneous, the cell potential E will be given by:
E = E°(reduction) - E°(oxidation)

E = E°(Cu2+ | Cu) - E°(I2 | I-)
E = 0.52 V - 0.54 V
E = -0.02 V

(b) If [Cu+] was equal to 1.2 M, at what concentration of I− would the cell have zero potential?

Let's consider the Nernst equation at zero potential, E = 0:

0 = E° - (RT/nF) * ln(Q)

At zero potential, the reaction quotient Q should be equal to 1. Let's substitute the given values and solve for [I-]:

E° = 0.52 V - 0.54 V = -0.02 V
R = 8.314 J/mol·K
T = 298 K
n = 2 (from the balanced half-reactions)
F = 96,485 C/mol
Q = [Cu+]/[I-]²

ln(Q) = (E° * nF) / (RT)
ln([Cu+]/[I-]²) = (-0.02 V * 2 * 96,485 C/mol) / (8.314 J/mol·K * 298 K)

Simplifying this equation will give us the solution for the concentration of [I-].

To determine the electromotive force (E) of the cell at the given concentrations, you can use the Nernst equation, which relates the concentration of reactants and products to the cell potential.

The Nernst equation is given as follows:
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
- n is the number of moles of electrons transferred in the balanced equation (in this case, both half-cell reactions transfer 1 electron)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which can be determined using the concentrations of the species involved in the redox reaction.

Let's solve part (a) first.

(a) To determine E for the cell at the given concentrations, we need to know the standard cell potential (E°) for the given half-cell reactions. Using a reliable reference, we can find that E° for the Cu+(aq)|Cu(s) half-cell reaction is +0.52 V, and E° for the I2(s)|I−(aq) half-cell reaction is +0.54 V.

Now, plug the values into the Nernst equation:
E = E° - (RT / nF) * ln(Q)

For the given half-cell reactions, we need to multiply the reduction half-reaction by its coefficient so that both reactions involve the same number of electrons. The balanced overall cell reaction is:

Cu+(aq) + I2(s) -> Cu(s) + 2 I−(aq)

The number of electrons transferred (n) is 1.

Since the concentration of Cu+(aq) and I−(aq) is 2.7 M, the reaction quotient (Q) equals the ratio of the product concentrations to the reactant concentrations raised to their stoichiometric coefficients (from the balanced equation):
Q = [Cu(s)]/[Cu+(aq)] * [I−(aq)]^2 / [I2(s)]

Substituting the given values into the equation:
Q = (2.7 M) / (2.7 M) * (2.7 M)^2 / (1 M)

Simplifying:
Q = (2.7 M)^2

Now, substitute all the known values into the Nernst equation:
E = E° - (RT / nF) * ln(Q)
E = +0.52 V - (8.314 J/(mol·K) * 298 K / (1 mol·1 F) * ln((2.7 M)^2)

E = +0.52 V - (8.314 J/(mol·K) * 298 K / (1 mol·1 F) * ln(7.29 M^2)

E = +0.52 V - (8.314 J/(mol·K) * 298 K / (1 mol·1 F) * ln(52.15 M)

E = +0.52 V - 0.0438 V

E = +0.4762 V (rounded to four significant figures)

Therefore, the cell potential at the given concentrations is approximately +0.4762 V.

Now let's solve part (b).

(b) To find the concentration of I− that would result in a zero cell potential when the [Cu+] is 1.2 M, we can set E equal to zero in the Nernst equation.

0 = E° - (RT / nF) * ln(Q)
0 = +0.52 V - (8.314 J/(mol·K) * 298 K / (1 mol·1 F) * ln((1.2 M) / [I−(aq)]^2)

Simplifying:
ln((1.2 M) / [I−(aq)]^2) = (0.52 V) / (8.314 J/(mol·K) * 298 K / (1 mol·1 F))

ln((1.2 M) / [I−(aq)]^2) = 0.0338 V

Now, we can solve for [I−(aq)].
[I−(aq)]^2 = (1.2 M) / e^(0.0338 V)

[I−(aq)] = sqrt((1.2 M) / e^(0.0338 V))

Substituting the values, solve for [I−(aq)].

[I−(aq)] = sqrt((1.2 M) / e^(0.0338 V))

[I−(aq)] = sqrt((1.2 M) / e^(0.0338 V))

[I−(aq)] = sqrt((1.2 M) / 1.0346)

[I−(aq)] = sqrt(1.1612 M)

[I−(aq)] = 1.077 M (rounded to three decimal places)

Therefore, if [Cu+] is equal to 1.2 M, the concentration of I− that would result in a zero cell potential is approximately 1.077 M.