Cell Potential Calculation using Nernst Equation

The question asks for the cell potential (\(E_{cell}\)) of the given electrochemical cell at 298 K and 1 bar pressure, considering the activity coefficient of the electrolyte.

The cell notation is Zn(s)|ZnBr₂(aq, 0.20 mol/kg) ||AgBr(s)|Ag(s)|Cu. We will consider the standard electrochemical cell reaction involving the Zn and AgBr/Ag couples. The presence of 'Cu' at the end of the notation seems extraneous to the cell reaction between the Zn and AgBr/Ag electrodes and will be ignored for the calculation of the cell potential using the Nernst equation for the Zn/AgBr system.

Identifying Half-Reactions

The cell notation indicates the anode (oxidation) on the left and the cathode (reduction) on the right:

• Anode (Oxidation): Zinc metal is oxidized to zinc ions.
• \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\)
• Cathode (Reduction): Silver bromide is reduced to silver metal and bromide ions.
• \(AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)\)

Balancing the Overall Reaction

To obtain the overall cell reaction, we multiply the cathode half-reaction by 2 to balance the number of electrons:

• Anode: \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\)
• Cathode: \(2 \times (AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq))\) which is \(2AgBr(s) + 2e^- \rightarrow 2Ag(s) + 2Br^-(aq)\)

Adding the two balanced half-reactions gives the overall reaction:

\(Zn(s) + 2AgBr(s) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2Br^-(aq)\)

From the balanced reaction, we see that \(n\), the number of electrons transferred, is 2.

Calculating Standard Cell Potential (\(E^0_{cell}\))

The standard cell potential is calculated as the difference between the standard reduction potentials of the cathode and the anode:

\(E^0_{cell} = E^0_{cathode} - E^0_{anode}\)

Given standard potentials (reduction potentials):

• \(E^0_{Zn^{+2}/Zn} = -0.762 \, V\)
• \(E^0_{AgBr/Ag} = +0.730 \, V\)

Here, the cathode is AgBr/Ag and the anode is Zn/Zn²⁺. So,

\(E^0_{cell} = E^0_{AgBr/Ag} - E^0_{Zn^{+2}/Zn} = (+0.730 \, V) - (-0.762 \, V) = 0.730 \, V + 0.762 \, V = 1.492 \, V\)

Applying the Nernst Equation

The cell potential at non-standard conditions is given by the Nernst equation. At 298 K, the Nernst equation is:

\(E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log_{10} Q\)

Where \(Q\) is the reaction quotient.

Calculating the Reaction Quotient (\(Q\))

For the overall reaction \(Zn(s) + 2AgBr(s) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2Br^-(aq)\), the reaction quotient \(Q\) is given by the activities of the products and reactants:

\(Q = \frac{a_{Zn^{2+}(aq)} \cdot a_{Ag(s)}^2 \cdot a_{Br^-(aq)}^2}{a_{Zn(s)} \cdot a_{AgBr(s)}^2}\)

The activities of pure solids (Zn(s), AgBr(s), Ag(s)) are considered to be 1. So,

\(Q = a_{Zn^{2+}(aq)} \cdot a_{Br^-(aq)}^2\)

The activities of the ions are related to their molalities (\(m\)) and activity coefficients (\(\gamma\)) by \(a_i = \gamma_i \frac{m_i}{m^0}\), where \(m^0\) is the standard molality (1 mol/kg). We are given the mean activity coefficient (\(\gamma_\pm\)) for the ZnBr₂ solution. A common approximation is to use the mean activity coefficient for individual ions: \(a_i \approx \gamma_\pm \frac{m_i}{m^0}\).

The molality of the ZnBr₂ solution is given as 0.20 mol/kg.

When ZnBr₂ dissolves, it dissociates as: \(ZnBr_2(aq) \rightarrow Zn^{2+}(aq) + 2Br^-(aq)\).

• Molality of Zn²⁺ ions, \(m_{Zn^{2+}} = 0.20\) mol/kg.
• Molality of Br^- ions, \(m_{Br^-} = 2 \times 0.20 = 0.40\) mol/kg.
• Mean activity coefficient, \(\gamma_\pm = 0.462\).
• Standard molality, \(m^0 = 1\) mol/kg.

Now, calculate the activities of the ions:

• Activity of Zn²⁺, \(a_{Zn^{2+}} = \gamma_\pm \frac{m_{Zn^{2+}}}{m^0} = 0.462 \times \frac{0.20}{1} = 0.0924\)
• Activity of Br^-, \(a_{Br^-} = \gamma_\pm \frac{m_{Br^-}}{m^0} = 0.462 \times \frac{0.40}{1} = 0.1848\)

Now, calculate the reaction quotient \(Q\):

\(Q = a_{Zn^{2+}} \cdot a_{Br^-}^2 = (0.0924) \cdot (0.1848)^2\)

\(Q = 0.0924 \cdot (0.03415104)\)

\(Q \approx 0.0031563\)

Calculating the Cell Potential (\(E_{cell}\))

Substitute the values into the Nernst equation (\(E^0_{cell} = 1.492 \, V\), \(n=2\), \(Q \approx 0.0031563\)):

\(E_{cell} = 1.492 \, V - \frac{0.0592}{2} \log_{10} (0.0031563)\)

\(E_{cell} = 1.492 \, V - 0.0296 \times \log_{10} (0.0031563)\)

Using a calculator, \(\log_{10} (0.0031563) \approx -2.5009\)

\(E_{cell} = 1.492 \, V - 0.0296 \times (-2.5009)\)

\(E_{cell} = 1.492 \, V + (0.0296 \times 2.5009)\)

\(E_{cell} = 1.492 \, V + 0.07402664 \, V\)

\(E_{cell} \approx 1.56602664 \, V\)

Rounding to three decimal places, the cell potential is 1.566 V.

Conclusion

The calculated cell potential at 298 K and 1 bar pressure for the given cell with the specified ZnBr₂ solution molality and mean activity coefficient is approximately 1.566 V.

The final answer is ≈ 1.566 V.