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 Change

In \( \ce{N2H4} \), let the oxidation state of \( \ce{N} \) be \( x \). Since \( \ce{H} \) is \( +1 \), we have \( 2x + 4(+1) = 0 \), so \( x = -2 \). 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

To balance \( \ce{H} \), add \( \ce{OH^-} \) and \( \ce{H2O} \). The number of \( \ce{H} \) atoms in \( \ce{N2H4} \) is 4. In basic solution, we can balance \( \ce{H} \) by adding \( 4\ce{OH^-} \) to the left and \( 4\ce{H2O} \) to the right or vice - versa. But first, let's balance the charge.

Step5: Balance Charge

The charge on the left side of \( \ce{N2H4 -> N2} \) (before balancing charge) is \( 0 \) (since \( \ce{N2H4} \) is neutral). The charge on the right side is \( 0 \) ( \( \ce{N2} \) is neutral). But we know from oxidation state change that each \( \ce{N} \) loses 2 electrons, 2 \( \ce{N} \) atoms lose a total of \( 4 \) electrons. So we add \( 4\ce{e^-} \) to the right side to balance the charge. The balanced half - reaction in basic solution (after considering charge and atom balance) is \( \ce{N2H4 + 4OH^- -> N2 + 4H2O + 4e^-} \). Let's check the \( \ce{H} \) and \( \ce{O} \) balance: Left side \( \ce{H} \): \( 4 + 4 = 8 \), right side \( \ce{H} \): \( 4\times2 = 8 \); Left side \( \ce{O} \): \( 4 \), right side \( \ce{O} \): \( 4 \). The number of electrons transferred is \( 4 \) (since 2 \( \ce{N} \) atoms, each losing 2 electrons, \( 2\times2 = 4 \)).

Answer:

The number of electrons transferred in the \( \ce{N2H4} \) half - reaction (oxidation half - reaction) is \( 4 \). The balanced half - reaction is \( \ce{N2H4 + 4OH^- -> N2 + 4H2O + 4e^-} \), and the number of electrons transferred is \( 4 \).