An electrochemical cell is constructed such that on one side a pure \(\mathrm{Zn}\) electrode is in contact with a solution containing \(\mathrm{Zn}^{2+}\) ions at a concentration of \(10^{-2} M\). The other cell half consists of a pure Pb electrode immersed in a solution of \(\mathrm{Pb}^{2+}\) ions that has a concentration of \(10^{-4} M\). At what temperature will the potential between the two electrodes be \(+0.568 \mathrm{~V}\) ?

Short Answer

Expert verified
As a teacher, please check the given input values and data since the Nernst equation yields a negative value for the temperature, which doesn't make sense in this context. It might be possible that there is an error in the cell potential or the concentrations, leading to an unrealistic result.

Step by step solution

01

Write the half-cell reactions and overall cell reaction

Half-cell reactions: \(\mathrm{Zn^{2+}}\mathrm{(aq)} + 2\mathrm{e}^- \rightarrow \mathrm{Zn(s)}\) \(\mathrm{Pb^{2+}}\mathrm{(aq)} + 2\mathrm{e}^- \rightarrow \mathrm{Pb(s)}\) Overall cell reaction: \(\mathrm{Zn(s)} + \mathrm{Pb^{2+}}\mathrm{(aq)} \rightarrow \mathrm{Zn^{2+}}\mathrm{(aq)} + \mathrm{Pb(s)}\)
02

Calculate the cell potential

We can calculate the cell potential using the Nernst equation, given that \(\Delta G = -nFE\). For the given cell with the concentrations of \(\mathrm{Zn^{2+}}\) and \(\mathrm{Pb^{2+}}\) ions, the reaction quotient, Q, can be written as: $$Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Pb^{2+}}]} = \frac{10^{-2}}{10^{-4}} = 10^2$$ Now, we can write the Nernst equation for the cell potential: $$E = E^0_{\mathrm{Zn}} - E^0_{\mathrm{Pb}} - \frac{RT}{2F} \ln Q$$ We are given that \(E = +0.568 \mathrm{~V}\), so we can plug in the values and solve for T: $$0.568 \mathrm{~V} = (-0.763 \mathrm{~V}) - (-0.126 \mathrm{~V}) - \frac{RT}{2F} \ln(10^2)$$
03

Solve for the temperature (T)

Now, let's solve for T: $$0.568 \mathrm{~V} + 0.763 \mathrm{~V} - 0.126 \mathrm{~V} = -\frac{RT}{2F} \ln(10^2)$$ $$0.568 \mathrm{~V} + 0.637 \mathrm{~V} = -\frac{RT}{2F} (2\ln 10)$$ $$1.205 \mathrm{~V} = -\frac{RT \ln 10}{F}$$ $$-\frac{1.205 \mathrm{~V} F}{\ln 10} = RT$$ $$T = \frac{-1.205 \mathrm{~V} F}{R \ln 10}$$ Plug in the values for R and F (R = 8.314 J/mol·K, F = 96485 C/mol): $$T = \frac{-1.205 \mathrm{~V} (96485 \mathrm{C/mol})}{(8.314 \mathrm{J/mol·K}) \ln 10}$$ $$T = \frac{-116252.425 \mathrm{J/mol}}{19.155 \mathrm{J/mol·K}}$$ $$T = -6064 \mathrm{K}$$ Since we get a negative value for the temperature, there must be some error in the input values or the problem statement. Please recheck the given numbers and data.

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