A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \operatorname{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 M,\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

Since the cell is composed of two silver-silver chloride half-cells with different chloride ion concentrations, the electrode with the higher concentration of chloride ions will serve as the anode, and the other electrode will serve as the cathode. In this case, the electrode with \(\left[\mathrm{Cl}^{-}\right]=2.55 M\) will be the anode, and the electrode with \(\left[\mathrm{Cl}^{-}\right]=0.0150 M\) will be the cathode.

The standard emf of the cell, \(E_{cell}^0\), can be calculated using the half-reaction standard reduction potential, \(E^0 = +0.222 \,\text{V}\) (found in a standard reduction potential table): $$ E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 $$ Since both electrodes involve the same half-reaction, we have: $$ E_{cell}^0 = E^0 - E^0 = 0 \,\text{V} $$

To calculate the cell emf for the given concentrations, we will use the Nernst equation: $$ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln{Q} $$ For this cell: - \(R = 8.314 \,\text{J/mol K}\) is the gas constant. - \(T = 298 \,\text{K}\) is the standard temperature. - \(n = 1\) is the number of moles of electrons transferred in the half-reaction. - \(F = 96,485 \,\text{C/mol}\) is the Faraday constant. - \(Q\) is the reaction quotient that can be calculated as \(\frac{[\mathrm{Cl}^{-}]_{anode}}{[\mathrm{Cl}^{-}]_{cathode}}\) in our case. Plugging in the values, we get: $$ E_{cell} = 0\, \text{V} - \frac{8.314 \,\text{J/mol K} \cdot 298 \,\text{K}}{(1)(96485 \, \text{C/mol})} \ln\left(\frac{2.55 \,\text{M}}{0.0150 \,\text{M}}\right) $$ Calculating the cell emf: $$ E_{cell} \approx +0.218 \,\text{V} $$

During the operation of the cell, the anode's half-reaction will occur spontaneously (AgCl will lose an electron and produce Ag and Cl-), while the cathode's half-reaction will occur non-spontaneously (Ag and Cl- combine to form AgCl and release an electron). Thus, - For the anode, since Ag and Cl- are being produced, \(\left[\mathrm{Cl}^{-}\right]\) will increase. - For the cathode, since AgCl is being formed, \(\left[\mathrm{Cl}^{-}\right]\) will decrease. In summary, the cathode has a chloride concentration of \(0.0150\, \mathrm{M}\), the standard emf of the cell is \(0 \,\text{V}\), the cell emf for the given concentrations is approximately \(+0.218\, \text{V}\), and the chloride concentration will increase at the anode and decrease at the cathode as the cell operates.