When 4 moles of electrons are added to NO_3^- | vedclass

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(B) The oxidation state of nitrogen in $NO_3^-$ is $+5$. When $4$ moles of electrons are added,the total change in oxidation state is $-4$. The final

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$1/2 \ mol \ N_2O$

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(B) The oxidation state of nitrogen in $NO_3^-$ is $+5$. When $4$ moles of electrons are added,the total change in oxidation state is $-4$. The final oxidation state of nitrogen will be $+5 - 4 = +1$. Among the options,$N_2O$ has nitrogen in the $+1$ oxidation state. The balanced half-reaction is: $NO_3^- + 4e^- + 2H^+ \longrightarrow \frac{1}{2} N_2O + \frac{5}{2} H_2O$. Thus,the product is $1/2 \ mol \ N_2O$.

When $4$ moles of electrons are added to $NO_3^-$,the product obtained is .......

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