The Ksp of Zn(OH)2 is 1.8 * 10^-14. Find Ecell for the half-reaction: Zn(OH)2(s) + 2 e- ⇌ Zn(s) + 2 OH-(aq)

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1. Identify the half-reaction given: \( \text{Zn(OH)}_2(s) + 2e^- \rightleftharpoons \text{Zn}(s) + 2\text{OH}^-(aq) \).
2. Use the Nernst equation to find the cell potential: \( E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \), where \( Q \) is the reaction quotient.
3. Determine \( E^\circ_{cell} \) using the standard reduction potentials. For the reaction \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s) \), find \( E^\circ \) from a standard reduction potential table.
4. Calculate the reaction quotient \( Q \) using the expression \( Q = \frac{[\text{OH}^-]^2}{1} \) since \( \text{Zn(OH)}_2 \) and \( \text{Zn} \) are solids and do not appear in the expression.
5. Substitute the values into the Nernst equation to solve for \( E_{cell} \). Remember to use \( n = 2 \) for the number of electrons transferred and the given \( K_{sp} \) to find \( [\text{OH}^-] \).