To calculate the equilibrium constant for the reaction provided and determine [tex]\(\Delta G^\circ\)[/tex], let's follow the step-by-step solution:
1. Write the half-reactions and their standard reduction potentials:
[tex]\[
\begin{array}{l}
Hg^{2+}(aq) + 2e^- \rightarrow Hg(l) \quad E_{red}^\circ = 0.855 \text{ V} \\
Ag^+(aq) + e^- \rightarrow Ag(s) \quad E_{red}^\circ = 0.799 \text{ V}
\end{array}
\][/tex]
2. Identify the anode and cathode reactions:
- The anode is where oxidation occurs: [tex]\(Ag(s) \rightarrow Ag^+(aq) + e^-\)[/tex]
- The cathode is where reduction occurs: [tex]\(Hg^{2+}(aq) + 2e^- \rightarrow Hg(l)\)[/tex]
3. Determine the standard cell potential [tex]\(E^\circ_{cell}\)[/tex]:
- Use the standard reduction potentials to find [tex]\(E^\circ_{cell}\)[/tex]:
[tex]\[
E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}
\][/tex]
- Here, [tex]\(E^\circ_{cathode} = 0.855 \text{ V}\)[/tex] and [tex]\(E^\circ_{anode} = 0.799 \text{ V}\)[/tex].
So,
[tex]\[
E^\circ_{cell} = 0.855 \text{ V} - 0.799 \text{ V} = 0.056 \text{ V}
\][/tex]
4. Calculate the equilibrium constant [tex]\(K\)[/tex] at 298 K:
- The relationship between [tex]\(E^\circ_{cell}\)[/tex] and the equilibrium constant [tex]\(K\)[/tex] is given by:
[tex]\[
E^\circ_{cell} = \frac{0.0591}{n} \log_{10}(K) \quad \text{(at 298 K)}
\][/tex]
- Here, [tex]\(n\)[/tex] is the number of moles of electrons transferred, which is 2.
Solving for [tex]\(\log_{10}(K)\)[/tex]:
[tex]\[
\log_{10}(K) = \frac{n \cdot E^\circ_{cell}}{0.0591}
\][/tex]
Plugging in the values:
[tex]\[
\log_{10}(K) = \frac{2 \cdot 0.056}{0.0591} = 1.8951
\][/tex]
- To find [tex]\(K\)[/tex], take the antilogarithm (base 10):
[tex]\[
K = 10^{1.8951} = 78.540
\][/tex]
5. Calculate [tex]\(\Delta G^\circ\)[/tex] for the reaction:
- Use the relationship between [tex]\(\Delta G^\circ\)[/tex], [tex]\(n\)[/tex], [tex]\(F\)[/tex], and [tex]\(E^\circ_{cell}\)[/tex]:
[tex]\[
\Delta G^\circ = -nFE^\circ_{cell}
\][/tex]
- Where [tex]\(F\)[/tex] is the Faraday constant, [tex]\(F = 96485 \text{ C/mol}\)[/tex].
Plugging in the values:
[tex]\[
\Delta G^\circ = -2 \cdot 96485 \cdot 0.056 = -10806.32 \text{ J/mol}
\][/tex]
6. Determine if [tex]\(\Delta G^\circ\)[/tex] is greater or less than zero:
- Since [tex]\(\Delta G^\circ = -10806.32 \text{ J/mol}\)[/tex], which is negative, [tex]\(\Delta G^\circ\)[/tex] for this reaction is less than zero.
Final Answers:
- The equilibrium constant [tex]\(K\)[/tex] at 298 K is: [tex]\(78.540\)[/tex].
- [tex]\(\Delta G^\circ\)[/tex] for this reaction would be less than zero.
