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?

To determine the cell potential (E) for the given 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
- 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 equation
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient

For the cell potential calculation, we need to find the standard cell potential (E°) for the given half-cell reactions. We can either look up the E° values in a reference table or calculate them using standard reduction potentials (E°red).

The standard reduction potential (E°red) for a half-cell reaction represents the potential of the reduction half-reaction at standard conditions (1 M concentration and 1 atm pressure). The E°red can be found in a table of standard reduction potentials.

Here are the E°red values for the given half-cell reactions:
Cu+(aq) + e- → Cu(s) : E°red = +0.52 V
I2(s) + 2e- → 2I-(aq) : E°red = +0.54 V

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

We'll first convert the half-cell reactions to their oxidation half-reactions and calculate E°cell.

Cu(s) → Cu+(aq) + e- : E°red = -0.52 V (reverse sign)
2I-(aq) → I2(s) + 2e- : E°red = +0.54 V

Since the cell is constructed by connecting the two half-cells, the E°cell can be calculated by summing the E°red values of the half-cells.

E°cell = E°red(cathode) - E°red(anode)
E°cell = (+0.54 V) - (-0.52 V) = +1.06 V

Now that we have E°cell, we can calculate E for the cell using the Nernst equation. Assuming room temperature (298 K), we'll substitute the values into the equation:

E = E° - (RT / nF) * ln(Q)
E = +1.06 V - ((8.314 J/(mol·K)) * (298 K) / (1 mol * 96485 C/mol)) * ln([I-]^2 / [Cu+] )
E = +1.06 V - (0.02569 V) * ln([I-]^2 / [Cu+])

Substitute the given concentration values: [Cu+] = 2.3 M and [I-] = 3.3 M

E = +1.06 V - (0.02569 V) * ln((3.3 M)^2 / (2.3 M))
E ≈ +1.06 V - (0.02569 V) * ln(4.53 M)

Using a calculator, calculate the natural logarithm (ln) of 4.53 and multiply it by -0.02569 V. Then subtract that from 1.06 V to find the value of E.

(b) If [Cu+] were equal to 1.0 M, at what concentration of I- would the cell have zero potential?

To find the concentration of I- at zero cell potential, we set E to zero in the Nernst equation and solve for the concentration of I-:

0 V = +1.06 V - (0.02569 V) * ln([I-]^2 / 1.0 M)

Using algebraic manipulation, isolate the ln term and solve for [I-]:

(0.02569 V) * ln([I-]^2 / 1.0 M) = +1.06 V

ln([I-]^2 / 1.0 M) = +1.06 V / (0.02569 V)

Using a calculator, divide 1.06 by 0.02569 to find the right side of the equation.

ln([I-]^2 / 1.0 M) ≈ +41.255

Next, we exponentiate both sides of the equation to eliminate the natural logarithm:

[I-]^2 / 1.0 M = e^(+41.255)

Using a calculator, exponentiate e by +41.255 to find the right side of the equation.

[I-]^2 / 1.0 M ≈ 1.69319321e+17

Now, multiply both sides of the equation by 1.0 M and take the square root to solve for [I-]:

[I-]^2 ≈ 1.69319321e+17 * 1.0 M

Using a calculator, multiply 1.69319321e+17 by 1.0 to find the right side of the equation.

[I-]^2 ≈ 1.69319321e+17 M

Finally, take the square root of both sides to find the concentration of I-:

[I-] ≈ sqrt(1.69319321e+17 M)

Using a calculator, take the square root of 1.69319321e+17 to find the concentration of I-.