A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\mathrm{Zn}\) electrodes. The two half-cells have \(\left[\mathrm{Zn}^{2+}\right]=1.8 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=\) \(1.00 \times 10^{-2} M\), respectively. (a) Which electrode is the anode 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 \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. Since both electrodes are made of Zinc, we will look at their respective concentrations to identify the anode and the cathode. The anode is the half-cell with the higher concentration, while the cathode is the half-cell with the lower concentration. In this case, the half-cell with \(\left[\mathrm{Zn}^{2+}\right]=1.8 M\) is the anode, and the half-cell with \(\left[\mathrm{Zn}^{2+}\right]=1.00 \times 10^{-2} M\) is the cathode.

The standard emf (electromotive force, also known as cell potential) of the cell can be determined using the standard reduction potentials of the half-cells. However, since both half-cells have the same redox couple (Zn²⁺/Zn), their standard reduction potentials are the same and cancel each other out. Thus, the standard emf of the cell is 0 V.

To calculate the cell emf for the given concentrations, we will use the Nernst equation: \[E = E^0 - \frac{RT}{nF} \ln Q\] Where: \(E\) is the cell potential \(E^0\) is the standard cell potential \(R\) is the gas constant (\(8.314 J K^{-1} mol^{-1}\)) \(T\) is the temperature (assumed to be 298 K) \(n\) is the number of electrons transferred in the half-cell reaction \(F\) is the Faraday constant (\(96485 C mol^{-1}\)) \(Q\) is the reaction quotient For the Zn/Zn²⁺ redox couple, the balanced half-reaction is: \[\mathrm{Zn^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Zn}\] The reaction quotient \(Q\) for this cell can be calculated as: \[Q = \frac{[\mathrm{Zn^{2+}}]_{cathode}}{[\mathrm{Zn^{2+}}]_{anode}} = \frac{1.00 \times 10^{-2} M}{1.8 M}\] Now, substitute all the values into the Nernst equation. Note that the standard cell potential, \(E^0\), is 0 V: \[E = 0 - \frac{8.314 J K^{-1} mol^{-1} \times 298 K}{2 \times 96485 C mol^{-1}} \ln \frac{1.00 \times 10^{-2} M}{1.8 M}\] Calculate the cell potential: \[E \approx -0.044 V\]

As the cell operates, oxidation will continue to occur at the anode, and reduction will occur at the cathode. Therefore: - At the anode, Zn will continue to oxidize to Zn²⁺, increasing the concentration of Zn²⁺. - At the cathode, Zn²⁺ will continue to reduce, forming Zn, thereby decreasing the concentration of Zn²⁺. In summary: a) The anode is the half-cell with \(\left[\mathrm{Zn}^{2+}\right]=1.8 M\). b) The standard emf of the cell is 0 V. c) The cell emf for the given concentrations is approximately -0.044 V. d) As the cell operates, the concentration of \(\left[\mathrm{Zn}^{2+}\right]\) will increase at the anode and decrease at the cathode.