Electrochemical cell potentials can be used to determine equilibrium constants that would be otherwise difficult to determine because concentrations are small.Calculate the value of K_sp for CdS from the following data.
CdS(s) + 2e F^- $\rightarrow$ Cd(s) +S² ^-(aq) $\quad$ $\quad$ E ° = -1.21 V
Cd² ⁺(aq) +2e^- $\rightarrow$ Cd(s) $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ E ° =-0.40 V

Question

Electrochemical cell potentials can be used to determine equilibrium constants that would be otherwise difficult to determine because concentrations are small.Calculate the value of K_sp for CdS from the following data.
CdS(s) + 2e F^-

\rightarrow

Cd(s) +S² ^-(aq)

\quad

\quad

E ° = -1.21 V
Cd² ⁺(aq) +2e^-

\rightarrow

Cd(s)

\quad

\quad

\quad

\quad

\quad

E ° =-0.40 V

Quizplus · Accepted Answer

The equilibrium constant $K_{sp}$ for CdS can be calculated using the Nernst equation and the given standard electrode potentials. The overall reaction for the dissolution of CdS in water is:$$CdS(s) \rightarrow Cd^{2+}(aq) + S^{2-}(aq)$$Given standard electrode potentials are:1. $E^\circ = 1.21\,V$ for the reduction of $Cd^{2+}$ to $Cd$.2. $E^\circ = 0.40\,V$ for the reduction of $S^{2-}$ to $S$.The overall cell potential $E^\circ_{cell}$ is the difference between the two potentials:$$E^\circ_{cell} = 1.21\,V - 0.40\,V = 0.81\,V$$The relationship between the standard cell potential and the equilibrium constant is given by the equation:$$E^\circ_{cell} = \frac{0.0592}{n} \log K$$For the dissolution of CdS, $n = 2$ (the number of electrons transferred in the overall reaction). Solving for $K$ gives:$$0.81 = \frac{0.0592}{2} \log K$$$$\log K = \frac{0.81 \times 2}{0.0592}$$$$\log K = 27.36$$Therefore, $K = 10^{27.36}$, which corresponds to choice B, $4.32 \times 10^{28}$.

Electrochemical Cell Potentials Can Be Used to Determine Equilibrium Constants →\rightarrow→

Electrochemical Cell Potentials Can Be Used to Determine Equilibrium Constants $\rightarrow$