For the balanced equation NO_3^- + 4H^+ + ne^ | vedclass

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(D) To find the value of $n$,we determine the change in the oxidation state of nitrogen in the reaction.In $NO_3^-$,the oxidation state of $N$ is $x +

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$3$

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(D) To find the value of $n$,we determine the change in the oxidation state of nitrogen in the reaction.In $NO_3^-$,the oxidation state of $N$ is $x + 3(-2) = -1$,so $x = +5$.In $NO$,the oxidation state of $N$ is $x + (-2) = 0$,so $x = +2$.The change in oxidation state is $|+5 - (+2)| = 3$.Since the oxidation state decreases by $3$,$3$ electrons are required for the reduction process.Therefore,$n = 3$.

For the balanced equation $NO_3^- + 4H^+ + ne^- \to 2H_2O + NO$,the value of $n$ is ........

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