Question

Calculate $E_{Cu^{+2}/Cu}$ at $pH=14$; given that $K_{sp}$ of $Cu(OH)_2 = 10^{-19}$; $E_{Cu^{+2}/Cu}^\circ = 0.034V$ at $298K$.

Answer 1

-0.22 V

Answer 2

To calculate the reduction potential $E_{Cu^{+2}/Cu}$ at pH=14, we need to determine the concentration of $Cu^{2+}$ ions under these conditions and then apply the Nernst equation.

1. Determine $[OH^-]$ from pH: Given pH = 14. At 298 K, pH + pOH = 14. So, pOH = 14 - pH = 14 - 14 = 0. Therefore, the concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^0 = 1$ M.

2. Determine $[Cu^{2+}]$ from $K_{sp}$: The dissolution equilibrium for copper(II) hydroxide is: $Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^-(aq)$ The solubility product constant ($K_{sp}$) is given by: $K_{sp} = [Cu^{2+}][OH^-]^2$ Given $K_{sp}$ of $Cu(OH)_2 = 10^{-19}$ and we found $[OH^-] = 1$ M. $10^{-19} = [Cu^{2+}](1)^2$ $[Cu^{2+}] = 10^{-19}$ M

3. Apply the Nernst Equation: The half-cell reaction for the copper couple is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$ The Nernst equation for this reaction at 298 K is: $E_{Cu^{2+}/Cu} = E^\circ_{Cu^{2+}/Cu} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$ Here, $n = 2$ (number of electrons transferred).

Note on $E^\circ_{Cu^{+2}/Cu}$: The question states $E^\circ_{Cu^{+2}/Cu} = 0.034V$. However, the standard reduction potential for $Cu^{2+}/Cu$ is commonly accepted as +0.34 V. The similar question also uses +0.34 V and leads to a standard result. Assuming the given value $0.034V$ is a typo and should be $0.34V$ to align with standard values and typical problem expectations:

Using $E^\circ_{Cu^{2+}/Cu} = 0.34$ V: $E_{Cu^{2+}/Cu} = 0.34 - \frac{0.059}{2} \log \frac{1}{10^{-19}}$ $E_{Cu^{2+}/Cu} = 0.34 - \frac{0.059}{2} \log (10^{19})$ $E_{Cu^{2+}/Cu} = 0.34 - (0.0295 \times 19)$ $E_{Cu^{2+}/Cu} = 0.34 - 0.5605$ $E_{Cu^{2+}/Cu} = -0.2205$ V

Rounding to two decimal places, $E_{Cu^{2+}/Cu} = -0.22$ V.

If we strictly use the value given in the question, $E^\circ_{Cu^{+2}/Cu} = 0.034V$: $E_{Cu^{2+}/Cu} = 0.034 - \frac{0.059}{2} \times 19$ $E_{Cu^{2+}/Cu} = 0.034 - 0.5605$ $E_{Cu^{2+}/Cu} = -0.5265$ V

Given the context of similar problems and standard values, it is highly probable that $E^\circ_{Cu^{+2}/Cu} = 0.34V$ was intended. Therefore, we proceed with the result obtained using $0.34V$.

The final answer is $\boxed{-0.22 V}$.

Explanation of the solution:

1. Calculate $[OH^-]$ from pH=14: pOH = 0, so $[OH^-] = 1$ M.
2. Calculate $[Cu^{2+}]$ from $K_{sp} = [Cu^{2+}][OH^-]^2$: $[Cu^{2+}] = K_{sp}/[OH^-]^2 = 10^{-19}/(1)^2 = 10^{-19}$ M.
3. Apply Nernst equation: $E_{Cu^{2+}/Cu} = E^\circ_{Cu^{2+}/Cu} - \frac{0.059}{2} \log \frac{1}{[Cu^{2+}]}$. Using $E^\circ_{Cu^{2+}/Cu} = 0.34 V$ (assuming typo correction from 0.034V): $E_{Cu^{2+}/Cu} = 0.34 - \frac{0.059}{2} \log(10^{19}) = 0.34 - 0.0295 \times 19 = 0.34 - 0.5605 = -0.2205 V$.

Answer:

The calculated $E_{Cu^{+2}/Cu}$ at pH=14 is -0.22 V.