Question

the two half-reactions are given below.
$2e^- + 2h_2o + hpo_3^{2-} \
ightarrow h_2po_2^- + 3oh^-$
$4oh^- + n_2h_4 \
ightarrow n_2 + 4h_2o + 4e^-$
what is the coefficient for the hydroxide ion in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in both half - reactions

The first half - reaction: \(2e^-+2H_2O + HPO_3^{2 - } ightarrow H_2PO_2^-+3OH^-\) has 2 electrons gained.
The second half - reaction: \(4OH^-+N_2H_4 ightarrow N_2 + 4H_2O+4e^-\) has 4 electrons lost.
To balance the number of electrons, we multiply the first half - reaction by 2.
After multiplying the first half - reaction by 2: \(4e^- + 4H_2O+2HPO_3^{2 - } ightarrow 2H_2PO_2^-+6OH^-\)

Step2: Add the two half - reactions together

Now we add the modified first half - reaction and the second half - reaction:
\((4e^- + 4H_2O+2HPO_3^{2 - })+(4OH^-+N_2H_4) ightarrow(2H_2PO_2^-+6OH^-)+(N_2 + 4H_2O+4e^-)\)

Step3: Cancel out common species

- Cancel out the \(4e^-\) on both sides.
- Cancel out \(4H_2O\) from both sides (4\(H_2O\) on the left and 4\(H_2O\) on the right).
After cancellation, we get: \(2HPO_3^{2 - }+N_2H_4 + 4OH^- ightarrow 2H_2PO_2^-+N_2+6OH^-\)

Step4: Simplify the hydroxide ions

Subtract 4\(OH^-\) from both sides: \(2HPO_3^{2 - }+N_2H_4 ightarrow 2H_2PO_2^-+N_2 + 2OH^-\)

Wait, let's re - check the addition:

First half - reaction (after multiplying by 2): \(4e^-+4H_2O + 2HPO_3^{2-} ightarrow2H_2PO_2^- + 6OH^-\)

Second half - reaction: \(4OH^-+N_2H_4 ightarrow N_2+4H_2O + 4e^-\)

Adding them:

Left side: \(4e^-+4H_2O + 2HPO_3^{2-}+4OH^-+N_2H_4\)

Right side: \(2H_2PO_2^- + 6OH^-+N_2+4H_2O + 4e^-\)

Cancel \(4e^-\) and \(4H_2O\):

Left: \(2HPO_3^{2-}+4OH^-+N_2H_4\)

Right: \(2H_2PO_2^- + 6OH^-+N_2\)

Now, subtract \(4OH^-\) from both sides:

Left: \(2HPO_3^{2-}+N_2H_4\)

Right: \(2H_2PO_2^- + 2OH^-+N_2\)

Wait, maybe I made a mistake in the first multiplication. Let's do it again.

First half - reaction: gain of 2 \(e^-\)

Second half - reaction: loss of 4 \(e^-\)

So multiply the first half - reaction by 2 to make electrons equal (2*2 = 4 electrons gained).

First half - reaction (multiplied by 2): \(4e^-+4H_2O+2HPO_3^{2 - } ightarrow2H_2PO_2^-+6OH^-\)

Second half - reaction: \(4OH^-+N_2H_4 ightarrow N_2+4H_2O + 4e^-\)

Now add:

\(4e^-+4H_2O + 2HPO_3^{2-}+4OH^-+N_2H_4=2H_2PO_2^-+6OH^-+N_2+4H_2O + 4e^-\)

Cancel \(4e^-\) and \(4H_2O\):

\(2HPO_3^{2-}+4OH^-+N_2H_4=2H_2PO_2^-+6OH^-+N_2\)

Now, subtract \(4OH^-\) from both sides:

\(2HPO_3^{2-}+N_2H_4=2H_2PO_2^-+2OH^-+N_2\)

Wait, no, the question is about the coefficient of \(OH^-\) in the combined balanced equation. Let's check the number of \(OH^-\) on the right and left after canceling.

Wait, another way:

First, find the number of electrons:

Reaction 1: reduction (gain 2 \(e^-\)): \(2e^-+2H_2O + HPO_3^{2-} ightarrow H_2PO_2^-+3OH^-\)

Reaction 2: oxidation (lose 4 \(e^-\)): \(4OH^-+N_2H_4 ightarrow N_2+4H_2O + 4e^-\)

To balance electrons, multiply reaction 1 by 2:

Reaction 1 (×2): \(4e^-+4H_2O + 2HPO_3^{2-} ightarrow 2H_2PO_2^-+6OH^-\)

Now add reaction 1 (×2) and reaction 2:

\(4e^-+4H_2O + 2HPO_3^{2-}+4OH^-+N_2H_4 ightarrow 2H_2PO_2^-+6OH^-+N_2+4H_2O + 4e^-\)

Now, cancel \(4e^-\), \(4H_2O\):

Left: \(2HPO_3^{2-}+4OH^-+N_2H_4\)

Right: \(2H_2PO_2^-+6OH^-+N_2\)

Now, the number of \(OH^-\) on the right is 6, on the left is 4. The net \(OH^-\) on the right is \(6 - 4=2\)? Wait, no, when we combine, we have to look at the final equation. Wait, maybe I messed up the direction. Let's do the addition again.

The left side has \(4OH^-\) and the right side has \(6OH^-\). So when we combine, the number of \(OH^-\) in the balanced equation is \(6 - 4 = 2\)? No, wait, let's write the overall equation:

After canceling electrons and water:

\(2HPO_3^{2-}+N_2H_4+4OH^- ightarrow2H_2PO_2^-+N_2+6OH^-\)

Now, subtract \(4OH^-\) from both sides:

\(2HPO_3^{2-}+N_2H_4 ightarrow2H_2PO_2^-+N_2 + 2OH^-\)

So the coefficient of \(OH^-\) is 2? Wait, no, let's check the electron transfer again.

Wait, the first half - reaction: H in \(HPO_3^{2-}\) (let's find oxidation state of H: let x be oxidation state of P, in \(HPO_3^{2-}\), \(1 + x+3\times(-2)=-2\), \(x = + 3\). In \(H_2PO_2^-\), let y be oxidation state of P, \(2\times(+1)+y + 2\times(-2)=-1\), \(y = + 1\). So P is reduced (from +3 to +1), gain of 2 electrons per P, so per \(HPO_3^{2-}\) (1 P) gain 2 electrons.

Second half - reaction: N in \(N_2H_4\) (oxidation state of N is - 2) to \(N_2\) (0), so each N loses 2 electrons, 2 N atoms, so total loss of 4 electrons.

So to balance electrons, multiply the reduction half - reaction (first one) by 2.

Reduction half - reaction (×2): \(4e^-+4H_2O + 2HPO_3^{2-} ightarrow 2H_2PO_2^-+6OH^-\)

Oxidation half - reaction: \(4OH^-+N_2H_4 ightarrow N_2+4H_2O + 4e^-\)

Now add them:

Left: \(4e^-+4H_2O + 2HPO_3^{2-}+4OH^-+N_2H_4\)

Right: \(2H_2PO_2^-+6OH^-+N_2+4H_2O + 4e^-\)

Cancel \(4e^-\) and \(4H_2O\):

Left: \(2HPO_3^{2-}+4OH^-+N_2H_4\)

Right: \(2H_2PO_2^-+6OH^-+N_2\)

Now, subtract \(4OH^-\) from both sides:

\(2HPO_3^{2-}+N_2H_4 ightarrow2H_2PO_2^-+N_2 + 2OH^-\)

Yes, so the coefficient of \(OH^-\) is 2. Wait, but let's check again.

Wait, maybe I made a mistake in the first multiplication. Let's do the addition without canceling water first.

Reduction (×2): \(4e^-+4H_2O + 2HPO_3^{2-} ightarrow2H_2PO_2^-+6OH^-\)

Oxidation: \(4OH^-+N_2H_4 ightarrow N_2+4H_2O + 4e^-\)

Adding:

\(4e^-+4H_2O + 2HPO_3^{2-}+4OH^-+N_2H_4=2H_2PO_2^-+6OH^-+N_2+4H_2O + 4e^-\)

Now, cancel \(4e^-\), \(4H_2O\):

\(2HPO_3^{2-}+4OH^-+N_2H_4=2H_2PO_2^-+6OH^-+N_2\)

Now, move \(4OH^-\) to the right:

\(2HPO_3^{2-}+N_2H_4=2H_2PO_2^-+N_2+6OH^--4OH^-\)

\(2HPO_3^{2-}+N_2H_4=2H_2PO_2^-+N_2 + 2OH^-\)

So the coefficient of \(OH^-\) is 2.

Answer:

2