The amount of electricity in Coulomb required for the oxidation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ is __________ $\times 10^5 \mathrm{C}$.
Answer (integer)
2
Solution
$\begin{aligned} & 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \\\\ & \frac{\mathrm{W}}{\mathrm{E}}=\frac{\mathrm{Q}}{96500} \\\\ & \text { mole } \times \text { n-factor }=\frac{\mathrm{Q}}{96500}\end{aligned}$
<br/><br/>$\begin{aligned} & 1 \times 2=\frac{Q}{96500} \\\\ & Q=2 \times 96500 \mathrm{C} \\\\ & =1.93 \times 10^5 \mathrm{C}\end{aligned}$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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