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?

ECu = Eo - 0.0592/1[log (Cu)/(Cu^+)] = ?

EI2 = Eo - 0.0592/2[log(I^-)^2/(I2)] = ?

Determine which is the more negative voltage and reverse that half cell and add it to the other one. Change the sign on Ecell for the reversed half cell and add E values to obtain Ecell.

b.
I would write the cell reaction and use
Ecell = Eocell - 0.0592/2(log Q) where
log Q = log(products)/(reactants)

To determine the cell potential (E) of a voltaic cell, we can use the Nernst equation:

E = E° - (RT/nF) * ln(Q)

Where:
- E is the cell potential
- E° is the standard cell potential (measured at 298 K with all species at 1 M concentration)
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (298 K in this case)
- n is the number of moles of electrons transferred in the balanced half-cell reaction
- F is Faraday's constant (96,485 C/mol)
- Q is the reaction quotient, which is the ratio of concentrations of products to reactants, each raised to the power of their stoichiometric coefficients.

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

For the given half-cell reactions:
Cu+(aq) + e− -> Cu(s) with E°1 = 0.52 V
I2(s) + 2 e− -> 2 I−(aq) with E°2 = 0.53 V

Since both half-cells are reduction reactions, the overall cell reaction is the sum of the half-cell reactions:

Cu+ (aq) + I2 (s) -> Cu (s) + 2 I- (aq)

The standard cell potential (E°) is the difference between the standard potentials of the reduction half-cells:

E°cell = E°red1 - E°red2
= 0.52 V - 0.53 V
= -0.01 V (rounded to two decimal places)

Now, we can use the Nernst equation to calculate the cell potential (E) at the given concentrations:

E = E° - (RT/nF) * ln(Q)

Where:
- R, T, and F are constants, so we need to determine the value of Q.

The concentration of Cu+ is 2.7 M, and the concentration of I- is also 2.7 M (as given in the question). The stoichiometric coefficients are 1 for Cu+ and 2 for I-. So, Q is calculated as:

Q = [I-]^2 / [Cu+]^1
= (2.7 M)^2 / (2.7 M)
= 2.7 M

Substituting the values into the Nernst equation:

E = -0.01 V - (8.314 J/mol·K * 298 K / 1 mol * 96,485 C/mol) * ln(2.7 M)
= -0.01 V - (8.314 J/mol·K * 298 K / 96,485 C) * ln(2.7)

Using proper units and rounding the result to the appropriate number of significant figures, we can determine the value of E.

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

To find the concentration of I- at which the cell potential is zero, we can set E = 0 in the Nernst equation and solve for Q.

0 = -0.01 V - (8.314 J/mol·K * 298 K / 1 mol * 96,485 C/mol) * ln(Q)

Rearranging the equation and solving for Q:

ln(Q) = (0.01 V) / (8.314 J/mol·K * 298 K / 1 mol * 96,485 C/mol)
Q = e^((0.01 V) / (8.314 J/mol·K * 298 K / 1 mol * 96,485 C/mol))

Next, we can use the concentration of Cu+ (1.2 M) and the stoichiometric coefficients (1 for Cu+ and 2 for I-) to determine the value of Q:

Q = [I-]^2 / [Cu+]^1
1.2 M = [I-]^2 / (1.2 M)
[I-]^2 = (1.2 M)^2
[I-] = √((1.2 M)^2)

By taking the square root of the result, we can determine the concentration of I- that would produce a zero cell potential.