A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With \(\left[\mathrm{Ag}^{+}\right]\) equal to $0.15 \mathrm{M}\(, at what concentration of \)\mathrm{I}^{-}$ would the cell have zero potential?

Step by step solution

01

Identifying the half-cell reactions

For each half-cell reaction, we can determine the standard reduction potentials. Using these we can find the overall cell reaction and calculate the standard cell potential, E°. Half-cell reactions: 1. \(Ag^+(aq) + 2e^- -> Ag(s)\) - Silver Half-cell 2. \(I_2(s) + 2e^- -> 2I^-(aq)\) - Iodine Half-cell

02

Calculate the standard cell potential (E°)

Access the standard reduction potentials for each half-cell reaction from a table. We find: 1. \(E°_{Ag} = +0.80V\) 2. \(E°_{I2} = +0.54V\) As the Silver half-cell has higher reduction potential, it is more likely to be reduced. Therefore, the Iodine half-cell will be oxidized and thus, our silver electrode will be the cathode, and our iodine electrode will be the anode. The standard cell potential can be calculated as follows: \(E°_{cell} = E°_{cathode} - E°_{anode} = E°_{Ag} - E°_{I2} = 0.80V - 0.54V = 0.26V\)

03

Apply the Nernst Equation to find the cell potential (E)

The Nernst Equation is given by \(E = E° - \frac{RT}{nF} \ln Q\), where \(E\) is the cell potential, \(E°\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of transferred electrons, \(F\) is the Faraday's constant, and \(Q\) is the reaction quotient. For our cell, we have \(T=298K\), \(n=2\), and \(R=8.314J.mol^{-1}.K^{-1}\), and \(F = 96,485 C.mol^{-1}\). We can also find Q using the concentrations given. Q can be calculated as follows: \(Q = \frac{[Ag^+]^2}{[I^-]^2}\). Now, substitute the known values into the Nernst Equation and solve for E: \(E = 0.26V - \frac{8.314*298}{2*96485} \ln{\frac{0.15^2}{0.035^2}}\)

04

Calculate cell potential (E)

On solving the Nernst Equation, we get the cell potential E: \(E \approx 0.357V\) Our final answers are: a) The cell potential at these concentrations is 0.357V. b) The anode of the cell is the iodine electrode. c) Yes, the answer for part (b) is the same as it would be if the cell were operated under standard conditions. d) To find the concentration of I⁻ when the cell has zero potential, set E = 0 in the Nernst Equation and solve for [I⁻]: \(0 = 0.26V - \frac{8.314*298}{2*96485} \ln{\frac{0.15^2}{[I^-]^2}}\) On solving this equation for [I⁻], we find: \([I^-] \approx 0.218M\) Therefore, with [Ag+] = 0.15M, the cell would have zero potential when [I⁻] = 0.218M.

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