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{hpo^2-_{3}} half - reaction. how many electrons are transferred?

Explanation:

Step1: Identify oxidation state changes

In \( \text{HPO}_3^{2-} \), let the oxidation state of P be \( x \). Hydrogen is +1, oxygen is -2. So \( 1 + x + 3(-2) = -2 \), solving gives \( x = +3 \). In \( \text{H}_2\text{PO}_2^- \), let P's oxidation state be \( y \). \( 2(+1) + y + 2(-2) = -1 \), solving gives \( y = +1 \). So P is reduced (oxidation state decreases from +3 to +1), so gain of electrons.

Step2: Balance atoms (except O and H)

P is already balanced: 1 P on each side.

Step3: Balance O with \( \text{H}_2\text{O} \)

Left side: 3 O (in \( \text{HPO}_3^{2-} \)), right side: 2 O (in \( \text{H}_2\text{PO}_2^- \)). So add 1 \( \text{H}_2\text{O} \) to the right:
\( \text{HPO}_3^{2-} ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} \)

Step4: Balance H with \( \text{H}^+ \) (then adjust for basic solution later)

Left side H: 1 (in \( \text{HPO}_3^{2-} \)), right side H: \( 2 + 2 = 4 \) (in \( \text{H}_2\text{PO}_2^- \) and \( \text{H}_2\text{O} \)). So add 3 \( \text{H}^+ \) to the left:
\( \text{HPO}_3^{2-} + 3\text{H}^+ ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} \)

Step5: Adjust for basic solution (add \( \text{OH}^- \) to both sides equal to \( \text{H}^+ \))

Add 3 \( \text{OH}^- \) to both sides:
\( \text{HPO}_3^{2-} + 3\text{H}^+ + 3\text{OH}^- ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \)
\( \text{H}^+ + \text{OH}^- = \text{H}_2\text{O} \), so left side becomes \( \text{HPO}_3^{2-} + 3\text{H}_2\text{O} \), right side becomes \( \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \). Simplify H₂O: subtract 1 H₂O from both sides:
\( \text{HPO}_3^{2-} + 2\text{H}_2\text{O} ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \) (wait, better way: after adding \( \text{OH}^- \), \( \text{H}^+ + \text{OH}^- = \text{H}_2\text{O} \), so 3 \( \text{H}^+ + 3\text{OH}^- = 3\text{H}_2\text{O} \). So equation becomes:
\( \text{HPO}_3^{2-} + 3\text{H}_2\text{O} ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \)
Simplify H₂O: \( 3\text{H}_2\text{O} - \text{H}_2\text{O} = 2\text{H}_2\text{O} \) on left? Wait, no: left side after adding 3 OH⁻: \( \text{HPO}_3^{2-} + 3\text{H}^+ + 3\text{OH}^- = \text{HPO}_3^{2-} + 3\text{H}_2\text{O} \). Right side: \( \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \). Wait, maybe I messed up. Let's redo:

Original after step 3: \( \text{HPO}_3^{2-} + 3\text{H}^+ ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} \)

Now, in basic solution, add \( \text{OH}^- \) to neutralize \( \text{H}^+ \). So add 3 \( \text{OH}^- \) to both sides:

\( \text{HPO}_3^{2-} + 3\text{H}^+ + 3\text{OH}^- ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \)

Left side: \( 3\text{H}^+ + 3\text{OH}^- = 3\text{H}_2\text{O} \), so:

\( \text{HPO}_3^{2-} + 3\text{H}_2\text{O} ightarrow \text{H}_2\text{PO}_2^- + \text{H}_2\text{O} + 3\text{OH}^- \)

Subtract \( \text{H}_2\text{O} \) from both sides:

\( \text{HPO}_3^{2-} + 2\text{H}_2\text{O} ightarrow \text{H}_2\text{PO}_2^- + 3\text{OH}^- \)

Now check O: left: 3 + 4 = 7? Wait no, \( \text{HPO}_3^{2-} \) has 3 O, \( 2\text{H}_2\text{O} \) has 2 O, total 5 O. Right: \( \text{H}_2\text{PO}_2^- \) has 2 O, \( 3\text{OH}^- \) has 3 O, total 5 O. Good. H: left: 1 + 4 = 5. Right: 2 + 3 = 5. Good. Charge: left: \( -2 + 0 = -2 \). Right: \( -1 + 3(-1) = -4 \). Wait, that's a problem. Wait, no, we missed electrons.

Step6: Balance charge with electrons

Left charge: \( -2 \) (from \( \text{HPO}_3^{2-} \)) + 0 (from \( 2\text{H}_2\text{O} \)) = \( -2 \). Right charge: \( -1 \) (from \( \text{H}_2\text{PO}_2^- \)) + 3(-1) (from \( 3\text{OH}^- \)) = \( -4 \). So we need to add electrons to the left (since reduction, gain electrons) to balance charge. The difference is \( -4 - (-2) = -2 \), so we need to add 2 electrons to the left (because electrons have negative charge, adding 2 e⁻ (charge -2) to left: \( -2 + (-2) = -4 \), which matches right side. Wait, no: let's do it properly.

The half-reaction is reduction (P gains electrons), so electrons are on the left (reactant side) or right? Wait, oxidation state of P: +3 to +1, so gain of 2 electrons (since each electron is -1, gain of 2 e⁻ means charge decreases by 2? Wait, no: oxidation state change is 3 - 1 = 2, so gain of 2 electrons (because reduction is gain of electrons). So the half-reaction should have 2 e⁻ on the left (reactant side) to balance charge.

Wait, let's redo the charge balance. After step 5 (before electrons), the equation is:

\( \text{HPO}_3^{2-} + 2\text{H}_2\text{O} ightarrow \text{H}_2\text{PO}_2^- + 3\text{OH}^- \)

Left charge: \( -2 + 0 = -2 \)

Right charge: \( -1 + 3(-1) = -4 \)

To balance charge, we need to add 2 electrons to the left (since electrons are negative, adding 2 e⁻ (charge -2) to left: \( -2 + (-2) = -4 \), which matches right side. So:

\( \text{HPO}_3^{2-} + 2\text{H}_2\text{O} + 2\text{e}^- ightarrow \text{H}_2\text{PO}_2^- + 3\text{OH}^- \)

Now check charge: left: -2 + 0 + (-2) = -4. Right: -1 + (-3) = -4. Good.

Now check oxidation state: P was +3, now +1, so gain of 2 electrons (since each electron is -1, gain of 2 e⁻ means oxidation state decreases by 2, which matches 3 - 1 = 2).

So the number of electrons transferred (gained) in this half-reaction is 2.

Answer:

2