Consider the following reaction. What is being oxidized?

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Answer 1 · 2017-09-12T15:41:00

Dihydrogen is being oxidized.....

Oxidation entails a formal LOSS of electrons, and an increase in oxidation number corresponding to the NUMBER of electrons lost...

We gots zerovalent dihydrogen gas, $stackrel(0)H_2$ , that is oxidized to water, i.e. $stackrel(+I)H$ :

$1/2H_2(g)rarrH^+ + 1e^-$ $(i)$

And since for every oxidation, every electron loss, there is a corresponding reduction, an electron gain....

$stackrel(+IV)CO_2(g)+2H^+ +2e^(-)rarrstackrel(+II)CO(g) +H_2O$ $(ii)$

And so we adds, $2xx(i)+(ii)$ .......to retire the electrons.....

$CO_2(g)+cancel(2H^+) +H_2(g) rarrCO(g) +H_2O+cancel(2H^+)$

To give....

$CO_2(g)+H_2(g) rarrCO(g) +H_2O$

And thus dihydrogen is oxidized and carbon dioxide is reduced......