Question

the redox reaction given below occurs in a basic solution. \\(\ce{hpo^{2-}_{3} + n_{2}h_{4} -> h_{2}po^{-}_{2} + n_{2}}\\) balance the \\(\ce{n_{2}h_{4}}\\) half - reaction. how many electrons are transferred?

Explanation:

Step1: Identify Oxidation State Changes

In \( \ce{N2H4} \), the oxidation state of \( \ce{N} \) is \(-2\) (since \( \ce{H} \) is \( +1 \), and \( 2x + 4(+1) = 0 \Rightarrow x = -2 \)). In \( \ce{N2} \), the oxidation state of \( \ce{N} \) is \( 0 \). Each \( \ce{N} \) atom loses \( 2 \) electrons (from \(-2\) to \( 0 \)), and there are \( 2 \) \( \ce{N} \) atoms in \( \ce{N2H4} \) and \( \ce{N2} \).

Step2: Write Unbalanced Half-Reaction

The oxidation half - reaction (since \( \ce{N} \) is oxidized) is: \( \ce{N2H4 -> N2} \)

Step3: Balance Atoms Other Than O and H

The number of \( \ce{N} \) atoms is already balanced (2 on each side).

Step4: Balance O (not needed here as no O atoms) and Balance H in Basic Solution

In basic solution, we add \( \ce{OH-} \) and \( \ce{H2O} \) to balance H. First, balance H by adding \( \ce{H2O} \) and \( \ce{OH-} \). The left - hand side has 4 H atoms. So we add \( 4\ce{OH-} \) to the right and \( 4\ce{H2O} \) to the left? Wait, no. The correct way: For the reaction \( \ce{N2H4 -> N2} \), to balance H, in basic solution, we can think of the reaction in terms of \( \ce{H2O} \) and \( \ce{OH-} \). The number of H atoms: left has 4 H. So we add \( 4\ce{H2O} \) to the right and \( 4\ce{OH-} \) to the left? Wait, no, the standard method for basic solution:

1. Balance H by adding \( \ce{H2O} \) to the side with less H and \( \ce{OH-} \) to the other side. The half - reaction \( \ce{N2H4 -> N2} \) has 4 H on the left. So we add \( 4\ce{H2O} \) to the right and \( 4\ce{OH-} \) to the left? Wait, no, let's do it step by step.

The unbalanced half - reaction (oxidation): \( \ce{N2H4 -> N2} \)

Balance H: Add \( 4\ce{H2O} \) to the right and \( 4\ce{OH-} \) to the left? No, the correct approach is:

The number of H atoms: left has 4 H. So we can write:

\( \ce{N2H4 + 4OH- -> N2 + 4H2O} \)

Now check H: left: 4 H (from \( \ce{N2H4} \)) + 4 H (from \( 4\ce{OH-} \))? No, wait, \( \ce{OH-} \) has one H. Wait, I made a mistake. Let's use the standard method for balancing redox half - reactions in basic solution:

1. Write the unbalanced half - reaction: \( \ce{N2H4 -> N2} \) (oxidation, since N is oxidized)
2. Balance N: already balanced (2 N on each side)
3. Balance H: There are 4 H on the left. In basic solution, we add \( \ce{H2O} \) to the side that needs H and \( \ce{OH-} \) to the other side. The formula for balancing H in basic solution is: if we have \( x \) H on the left, we can add \( x\ce{H2O} \) to the left and \( x\ce{OH-} \) to the right, or vice - versa. Wait, the correct way is:

For the reaction \( \ce{N2H4 -> N2} \), the number of H atoms: left has 4 H. So we add \( 4\ce{H2O} \) to the right and \( 4\ce{OH-} \) to the left? No, let's use the oxidation state to find the electrons.

From the oxidation state change: each N atom goes from - 2 to 0, so each N loses 2 electrons. There are 2 N atoms in \( \ce{N2H4} \), so total electrons lost: \( 2\times2 = 4 \) electrons.

Now, let's balance the half - reaction properly:

Oxidation half - reaction: \( \ce{N2H4 -> N2} \)

Balance H: Add \( 4\ce{H2O} \) to the right and \( 4\ce{OH-} \) to the left? Wait, no, let's do it by adding \( \ce{OH-} \) and \( \ce{H2O} \) correctly.

The number of H atoms: left has 4 H. So we can write:

\( \ce{N2H4 + 4OH- -> N2 + 4H2O} \)

Now check the charge:

Left - hand side charge: charge of \( \ce{N2H4} \) is 0, charge of \( 4\ce{OH-} \) is - 4, total charge: - 4.

Right - hand side charge: charge of \( \ce{N2} \) is 0, charge of \( 4\ce{H2O} \) is 0, total charge: 0. Wait, that's not balanced. We need to balance the charge by adding electrons.

The charge on the left is - 4, on the right is 0. Since it's an oxidation reaction (losing electrons), we add electrons to the right. The difference in charge is \( 0-(-4)= + 4 \), but since we are losing electrons (oxidation), the number of electrons lost is equal to the positive charge we need to add to the right. Wait, no: the oxidation half - reaction involves loss of electrons. So the charge on the left: \( \ce{N2H4} \) is neutral, \( 4\ce{OH-} \) is - 4, total - 4. The charge on the right: \( \ce{N2} \) is neutral, \( 4\ce{H2O} \) is neutral. So to balance the charge, we need to have a charge of - 4 on the right as well? No, that's wrong. Let's go back to the oxidation state.

Each N atom in \( \ce{N2H4} \) has an oxidation state of - 2, and in \( \ce{N2} \) it's 0. So for 2 N atoms, the change in oxidation state is \( 2\times(0 - (-2))=4 \). So the number of electrons lost is 4. So the balanced half - reaction (including electrons) is:

\( \ce{N2H4 + 4OH- -> N2 + 4H2O + 4e-} \)

Let's check the charge:

Left: \( \ce{N2H4} \) (0) + \( 4\ce{OH-} \) (- 4) total charge: - 4

Right: \( \ce{N2} \) (0) + \( 4\ce{H2O} \) (0) + \( 4e- \) (- 4) total charge: - 4

Atoms:

N: 2 on left, 2 on right.

H: 4 (from \( \ce{N2H4} \)) + 4 (from \( 4\ce{OH-} \))? No, wait \( \ce{OH-} \) has 1 H each, 4 \( \ce{OH-} \) has 4 H, and \( \ce{N2H4} \) has 4 H, total 8 H on left? No, I made a mistake. Wait \( \ce{N2H4} \) has 4 H atoms. The correct way to balance H in basic solution for \( \ce{N2H4 -> N2} \):

We can think of the reaction as \( \ce{N2H4 -> N2 + 4H+} \) (in acidic solution), but in basic solution, we add \( 4\ce{OH-} \) to both sides to neutralize \( \ce{H+} \):

\( \ce{N2H4 + 4OH- -> N2 + 4H2O} \)

Now, the oxidation state change: each N goes from - 2 to 0, 2 N atoms, so total loss of \( 2\times2 = 4 \) electrons. So the balanced half - reaction with electrons is \( \ce{N2H4 + 4OH- -> N2 + 4H2O + 4e-} \)

Answer:

The number of electrons transferred in the \( \ce{N2H4} \) half - reaction (oxidation half - reaction) is 4.