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Calculate the equilibrium constant for the following electrochemical reaction at a temperature of 25°C:$$\ce{Co^{2+} + 2e^- <=> Co(s)}$$Given that the standard reduction potential for the half-reaction is $-0.28\,\text{V}$ and the concentration of $\ce{Co^{2+}}$ is $0.1\,\text{M}$.

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To calculate the equilibrium constant  K  for the given electrochemical reaction, we can use the Nernst equation. The Nernst equation relates the reduction potential of a half-cell  E  to the standard reduction potential  E , temperature  T , and concentrations of the species involved in the reaction. The equation is:$$E = E - \frac{RT}{nF} \ln Q$$Where:- E is the reduction potential of the half-cell- E is the standard reduction potential  -0.28 V in this case - R is the gas constant  8.314 J/molK - T is the temperature in Kelvin  25C = 298.15 K - n is the number of electrons transferred in the half-reaction  2 in this case - F is the Faraday constant  96485 C/mol - Q is the reaction quotient, which is equal to the equilibrium constant  K  at equilibriumAt equilibrium, the reduction potential  E  is zero. Therefore, we can rearrange the Nernst equation to solve for K:$$0 = -0.28 - \frac{8.314 \times 298.15}{2 \times 96485} \ln K$$Solving for K:$$\ln K = \frac{-0.28 \times 2 \times 96485}{8.314 \times 298.15}$$$$K = e^{\frac{-0.28 \times 2 \times 96485}{8.314 \times 298.15}}$$$$K \approx 1.1 \times 10^6$$So, the equilibrium constant for the given electrochemical reaction is approximately $1.1 \times 10^6$.

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