Electrochemistry Question 2
Question 2 - 2024 (01 Feb Shift 2)
The amount of electricity in Coulomb required for the oxidation of $1 \mathrm{~mol}^{2}$ of $\mathrm{H}{2} \mathrm{O}$ to $\mathrm{O}{2}$ is _________ $\times 10^{5} \mathrm{C}$.
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Answer (2)
Solution
$2 \mathrm{H}{2} \mathrm{O} \rightarrow \mathrm{O}{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-}$
$\frac{W}{E}=\frac{Q}{96500}$
mole $\times$ n-factor $=\frac{\mathrm{Q}}{96500}$
$1 \times 2=\frac{Q}{96500}$
$Q=2 \times 96500 \mathrm{C}$
$=1.93 \times 10^{5} \mathrm{C}$
