Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ As the reaction progresses, what will happen to the overall voltage of the cell? (A) It will increase as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (B) It will increase as \(\left[\mathrm{Cu}^{+}\right]\) increases. (C) It will decrease as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (D) The voltage will remain constant.

Step by step solution

01

Analyze the half-cells and the cell reaction

In the given set-up, two half-cells are set up with half-cell A having a strip of copper (\(\mathrm{Cu}(s)\)) in copper nitrate (\(\mathrm{CuNO}_{3}(a q)\)), and half-cell B having a strip of zinc (\(\mathrm{Zn}(s)\)) in zinc nitrate (\(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq)). Based on the diagram, the given cell reaction is: \(2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q)\) with an electromotive force (E) of +1.28 V.

02

Understand the effect of changing concentrations on cell potential

According to the Nernst equation, the cell potential or voltage is affected by the concentrations of the species in the reaction. The Nernst equation leads to the cell potential equal to the standard cell potential minus a term that involves the log of the ratio of the product and reactant concentrations. However, in the given reaction, the cells have been connected like in a standard cell with Cu electrode as the cathode and Zn electrode as the anode. This means, here E(cell) = E(cathode) - E(anode). The standard cell potential for Cu is +0.34V and for Zn is -0.76V. Substituting these values, we get E(cell) = 0.34 - (-0.76) = 1.10V. But the given reaction shows that E(cell) is 1.28V. This shows that our metallic ions are not in their standard state of 1 M concentration.

03

Determine the effect of the progression of the reaction on the cell voltage

As the reaction progresses, the concentration of the metallic ions in the solution changes. The reaction uses up copper ions (\(Cu^{+}\)) and produces zinc ions (\(Zn^{2+}\)), so that the concentration of \(Cu^{+}\) decreases and that of \(Zn^{2+}\) increases. How these changes affect the cell voltage can be determined by looking at the Nernst equation again and understanding that increasing reactant concentration or decreasing product concentration would increase the cell potential. But here, as with progressing reaction our product \(Zn^{2+}\) concentration increases and \([Cu^{+}]\) decreases, it leads to a decrease in cell potential.

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