Question

Is ${\mathrm{H}}_2\mathrm{Se}$ reducing agent?

Answer 1

Reducing agent:

1. In a redox chemical reaction, the reducing agent loses one or more of its electrons to the electron receiving group.
2. After losing electrons, the reducing agent becomes oxidized and, by giving electrons, also causes the opposite reactant to become reduced.
3. When ${\mathrm{H}}_2\mathrm{Se}$ reacts with zinc we get zinc selenide with hydrogen gas.$\underset{\mathrm{Hydrogen}\mathrm{selenide}}{\stackrel{+1}{{\mathrm{H}}_2}\mathrm{Se}\left(\mathrm{s}\right)}+\stackrel{0}{\mathrm{Zn}}\to \underset{\mathrm{Zinc}\mathrm{selenide}}{\stackrel{+2}{\mathrm{Zn}}\mathrm{Se}\left(\mathrm{s}\right)}+\underset{\mathrm{hydrogen}\mathrm{gas}}{\stackrel{0}{{\mathrm{H}}_2}\left(\mathrm{g}\right)}$
4. From the reaction we can see that Zinc is oxidized but ${\mathrm{H}}_2$ is reduced, ${\mathrm{H}}_2\mathrm{Se}$ is the oxidizing agent and not the reducing agent.

Therefore, ${\mathrm{H}}_2\mathrm{Se}$ is not a reducing agent.