Question

question 7 (4 points)
balance the following half - reaction in an acidic solution:
hno$_{2(aq)}$ → no$_{3}^{-}$(aq)

Explanation:

Step1: Determine oxidation states

The oxidation state of N in $HNO_2$ is +3 and in $NO_3^-$ is +5. The change in oxidation state is +2, so 2 electrons are lost.

Step2: Balance atoms other than H and O

N atoms are already balanced as there is 1 N on each side.

Step3: Balance O atoms

There are 2 O in $HNO_2$ and 3 O in $NO_3^-$. Add 1 $H_2O$ to the left - hand side to balance O atoms: $HNO_2(aq)+H_2O(l)\to NO_3^-(aq)$.

Step4: Balance H atoms

There are 3 H on the left - hand side. Add 3 $H^+$ to the right - hand side to balance H atoms: $HNO_2(aq)+H_2O(l)\to NO_3^-(aq)+3H^+(aq)$.

Step5: Balance charge

The left - hand side has a net charge of 0, and the right - hand side has a net charge of $3+( - 1)=+2$. Add 2 electrons to the right - hand side to balance the charge: $HNO_2(aq)+H_2O(l)\to NO_3^-(aq)+3H^+(aq)+2e^-$

Answer:

$HNO_2(aq)+H_2O(l)\to NO_3^-(aq)+3H^+(aq)+2e^-$