Question

question 7 (5 points)
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during the electrolysis of an aqueous solution of sodium nitrate, a gas forms at the anode, what gas is it?
a) hydrogen
b) water
c) sodium
d) oxygen

Explanation:

In the electrolysis of aqueous sodium nitrate ($\ce{NaNO3}$), the ions present are $\ce{Na+}$, $\ce{NO3-}$, and $\ce{H2O}$ (which dissociates into $\ce{H+}$ and $\ce{OH-}$). At the anode (oxidation site), the possible species to be oxidized are $\ce{OH-}$ (from water) or $\ce{NO3-}$. The oxidation of $\ce{OH-}$ (or water) is easier than that of $\ce{NO3-}$. The reaction for the oxidation of $\ce{OH-}$ is $4\ce{OH-}
ightarrow \ce{O2} + 2\ce{H2O} + 4e^-$ (or from water: $2\ce{H2O}
ightarrow \ce{O2} + 4\ce{H+} + 4e^-$), producing oxygen gas. Hydrogen is formed at the cathode (reduction), water is not a gas in this context, and sodium is a metal that would not form at the anode in aqueous solution (as $\ce{Na+}$ is harder to reduce than $\ce{H+}$). So the gas at the anode is oxygen.

Answer:

D) Oxygen