91Ó°ÊÓ

First, we need to combine the two given half-reactions into one balanced cell reaction. To do this, we need to add the two half-reactions together, making sure to multiply them by appropriate coefficients to balance the electrons. The balanced cell reaction is: $$2\mathrm{Ag}^{+}+\mathrm{H}_{2}\mathrm{O}_{2}+2\mathrm{H}^{+}\longrightarrow 2\mathrm{Ag}+2\mathrm{H}_{2}\mathrm{O}$$

To find the standard cell potential, we will need to subtract the standard reduction potential of the reduction half-reaction (the first reaction) from the standard reduction potential of the oxidation half-reaction (the second reaction). This information is normally provided or can be looked up in tables. Because it is not given in this exercise, we will denote the standard cell potential \(\mathscr{E}^{\circ}_{\text{cell}}\) as a variable.

Now that we have the balanced cell reaction, we can apply the Nernst equation to calculate \(\mathscr{E}_{\text{cell}}\) for both scenarios (a and b). The Nernst equation is given as: $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln\left(\frac{\text{products}}{\text{reactants}}\right)$$ where \(R\) is the universal gas constant (\(8.314\ \text{J/mol}\cdot\text{K}\)), \(T\) is the temperature in Kelvin, \(n\) is the number of transferred electrons, and \(F\) is the Faraday constant (\(96485\ \text{C/mol}\)). For both scenarios (a and b), the temperature is given as \(25^{\circ}\mathrm{C}\) which equals to \(298.15\ \text{K}\), and the number of transferred electrons is 2 (from the balanced cell reaction). Scenario a:

In scenario a, we have following concentrations: \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\left[\mathrm{H}_{2}\mathrm{O}_{2}\right]=2.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) Using these concentrations, we can plug in the values into the Nernst equation: $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{\left[\mathrm{H}_{2}\mathrm{O}\right]^2}{\left[\mathrm{Ag}^{+}\right]^2 \left[\mathrm{H}_{2}\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{1}{(1.0)^2(2.0)(2.0)^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(0.25)$$ Scenario b:

In scenario b, we have following concentrations: \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2}\mathrm{O}_{2}\right]=1.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\ \mathrm{M}\) Using these concentrations, we can plug in the values into the Nernst equation: $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{\left[\mathrm{H}_{2}\mathrm{O}\right]^2}{\left[\mathrm{Ag}^{+}\right]^2 \left[\mathrm{H}_{2}\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - \frac{(8.314)(298.15)}{(2)(96485)} \ln\left(\frac{1}{(2.0)^2(1.0)(1.0 \times 10^{-7})^2}\right)$$ $$\mathscr{E}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(5\times10^{14})$$

Now that we have found the cell potentials (\(\mathscr{E}_{\text{cell}}\)) for both scenarios, we can compare them to the standard cell potential (\(\mathscr{E}^{\circ}_{\text{cell}}\)). Scenario a: If \(\mathscr{E}_{\text{cell}}>\mathscr{E}^{\circ}_{\text{cell}}\), then \(\mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(0.25)>\mathscr{E}^{\circ}_{\text{cell}}\). As the \(\ln(0.25)=-1.386\), the term \(-0.0257\ \ln(0.25)=0.03565\) is positive. It is clear that \(\mathscr{E}_{\text{cell}} > \mathscr{E}^{\circ}_{\text{cell}}\) for scenario a. Scenario b: If \(\mathscr{E}_{\text{cell}}<\mathscr{E}^{\circ}_{\text{cell}}\), then \(\mathscr{E}^{\circ}_{\text{cell}} - 0.0257\ \ln(5\times10^{14})<\mathscr{E}^{\circ}_{\text{cell}}\). As the \(\ln(5\times10^{14})=33.835\), the term \(-0.0257\ \ln(5\times10^{14})=-0.86837\) is negative. It is clear that \(\mathscr{E}_{\text{cell}} < \mathscr{E}^{\circ}_{\text{cell}}\) for scenario b. So, the cell potential is larger than the standard cell potential for scenario a, while the cell potential is smaller than the standard cell potential for scenario b.