The $K_{ep}$ of $Zn(OH{)}_2$ is $1\mathrm{.}8{10}^{-14}.$ For the $Z\frac{n^{2+}}{Z}n$ hall$-$reaction:

$Z{n}^{2+}(aq)+2e^{-}Zn(s),E=-0\mathrm{.}76V$

You can consider the reduction of $Zn(OH{)}_2$ as a two$-$step process that involves the dissociation of the hydroxide with the formation of $Z{n}^{2+}$ cations and the reduction of $Z{n}^{2+}$ :

$Zn(OH{)}_2(s)Z{n}^{2+}(aq)+2O{H}^{-}(aq)$

$Z{n}^{2+}(aq)+2e^{-}Zn(s)$

Calculate the electrode potential $E$ for the following half$-$reaction.

$Zn(OH{)}_2(s)+2e^{-}Zn(s)+2O{H}^{-}(aq)$

Assume the hall$-$reaction occurs in pure water. Since only the reduction of $Z{n}^{2+}$ contributes to the electrode potentlal, calculate the electrode potential for the initial half$-$reaction using the Nernst equation for the $Z\frac{n^{2+}}{Z}n$ hall$-$reaction.

Express your answer in volts to two decimal places.

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