Question

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Explanation:

1. First, balance the atoms other than O and H in the half - reaction \( \text{NO}_3^-

ightarrow \text{NO}_2 \). The number of N atoms is already balanced (1 N on each side).

1. Balance O atoms by adding \( \text{H}_2\text{O} \). There are 3 O in \( \text{NO}_3^- \) and 2 O in \( \text{NO}_2 \), so we add 1 \( \text{H}_2\text{O} \) to the product side: \( \text{NO}_3^-

ightarrow \text{NO}_2+\text{H}_2\text{O} \).

1. Balance H atoms by adding \( \text{H}^+ \) (since it's acidic solution). There are 2 H in \( \text{H}_2\text{O} \), so we add 2 \( \text{H}^+ \) to the reactant side: \( \text{NO}_3^- + 2\text{H}^+

ightarrow \text{NO}_2+\text{H}_2\text{O} \).

1. Now balance the charge. The charge on the left side: \( \text{NO}_3^- \) has a charge of - 1 and \( 2\text{H}^+ \) has a charge of + 2, so total charge is \( - 1+2 = + 1 \). The charge on the right side: \( \text{NO}_2 \) is neutral and \( \text{H}_2\text{O} \) is neutral, so total charge is 0. To balance the charge, we need to add 1 electron to the left side (reactant side) because we need to decrease the positive charge on the left. The electron - addition for charge balance: \( \text{NO}_3^- + 2\text{H}^++e^-

ightarrow \text{NO}_2+\text{H}_2\text{O} \). So electrons are on the reactant side.

Answer:

A