Question

how many water molecules should be added to the reactant side?
?h₂o + mno₄⁻ → mno₂ + 4oh⁻

Explanation:

Step1: Analyze oxygen atoms

On the product side, we have \(MnO_2\) (2 O atoms) and \(4OH^-\) (4 O atoms), so total O atoms on product side: \(2 + 4=6\). On the reactant side, we have \(MnO_4^-\) (4 O atoms) and \(xH_2O\) (x O atoms, where x is the number of \(H_2O\) molecules). So \(4 + x=6\), which gives \(x = 2\) initially? Wait, no, let's check hydrogen.

Step2: Analyze hydrogen atoms

On the product side, \(4OH^-\) has 4 H atoms. On the reactant side, \(xH_2O\) has \(2x\) H atoms. So \(2x = 4\), which gives \(x = 2\)? Wait, but let's check charge and Mn. Wait, Mn in \(MnO_4^-\) is +7, in \(MnO_2\) is +4, so gain of 3 electrons. But for balancing, let's do atom balance.

Wait, let's balance step by step:

1. Balance Mn: 1 \(MnO_4^-\) and 1 \(MnO_2\), so Mn is balanced.

2. Balance O: \(MnO_4^-\) has 4 O, \(MnO_2\) has 2 O, \(4OH^-\) has 4 O. So total O on product: \(2 + 4 = 6\). On reactant: \(MnO_4^-\) (4 O) + \(xH_2O\) (x O). So \(4 + x=6\) ⇒ \(x = 2\)? But then H: \(xH_2O\) has \(2x\) H, product has \(4OH^-\) (4 H). So \(2x = 4\) ⇒ \(x = 2\). Wait, but let's check charge.

Charge on reactant: \(MnO_4^-\) is -1, \(H_2O\) is neutral, so total charge: -1.

Charge on product: \(MnO_2\) is neutral, \(4OH^-\) is -4, so total charge: -4. Wait, that's a problem. Wait, maybe I made a mistake. Wait, let's do redox balancing properly.

Wait, Mn: +7 in \(MnO_4^-\) to +4 in \(MnO_2\): gain 3 e⁻.

Oxygen: Let's balance O by adding \(H_2O\) and \(OH^-\). Wait, the reaction is in basic medium? Wait, product has \(OH^-\), so basic.

Wait, another approach: balance H and O by \(H_2O\) and \(OH^-\).

On product side, \(4OH^-\) means 4 H and 4 O (from OH⁻) plus 2 O from \(MnO_2\), total O: 6, H: 4.

On reactant side, \(MnO_4^-\) has 4 O, so we need 2 more O (from \(H_2O\))? Wait, no, \(H_2O\) has H and O. Wait, the number of H on product is 4 (from 4 \(OH^-\)), so on reactant, we need 2 \(H_2O\) (since 2 \(H_2O\) has 4 H). Let's check O: 2 \(H_2O\) has 2 O, plus \(MnO_4^-\) has 4 O, total O: 6. Product: \(MnO_2\) (2 O) + 4 \(OH^-\) (4 O) = 6 O. Perfect.

Wait, but let's check charge. Reactant: \(MnO_4^-\) (-1) + 2 \(H_2O\) (0) = -1. Product: \(MnO_2\) (0) + 4 \(OH^-\) (-4) = -4. Wait, that's not balanced. Oh, I forgot to balance charge. Wait, Mn gains 3 e⁻, so we need to balance charge. Wait, maybe I missed electrons. Wait, no, the question is about balancing atoms (H and O) for \(H_2O\) in a redox reaction, but maybe the question is just about atom balance (H and O) since it's asking for \(H_2O\) molecules, so focusing on H and O.

Since product has 4 H (from 4 \(OH^-\)), reactant \(H_2O\) must provide 4 H, so number of \(H_2O\) is 2 (since each \(H_2O\) has 2 H, 2*2=4 H). And O: 2 \(H_2O\) has 2 O, plus \(MnO_4^-\) has 4 O, total 6 O. Product: \(MnO_2\) (2 O) + 4 \(OH^-\) (4 O) = 6 O. Perfect. Wait, but earlier charge was unbalanced, but maybe the question is just about atom balance for \(H_2O\) to balance H and O, ignoring charge for the purpose of finding \(H_2O\) molecules? Wait, no, let's do it again.

Wait, the correct way: balance H and O first.

Number of H on product: 4 (from 4 \(OH^-\)). So on reactant, \(H_2O\) must have 4 H, so 2 \(H_2O\) (since 2*2=4 H). Then O: 2 \(H_2O\) has 2 O, \(MnO_4^-\) has 4 O, total 6 O. Product: \(MnO_2\) (2 O) + 4 \(OH^-\) (4 O) = 6 O. Perfect. So the number of \(H_2O\) is 2? Wait, no, wait the answer is 3? Wait, no, let's check again.

Wait, maybe I made a mistake. Wait, let's write the reaction:

\(xH_2O + MnO_4^- ightarrow MnO_2 + 4OH^-\)

Balance H: left side has \(2x\) H, right side has 4 H (from 4 \(OH^-\)). So \(2x = 4\) ⇒ \(x = 2\).

Balance O: left side has \(x + 4\) O (x from \(H_2O\), 4 from \(MnO_4^-\)), right side has \(2 + 4 = 6\) O (2 from \(MnO_2\), 4 from \(OH^-\)). So \(x + 4 = 6\) ⇒ \(x = 2\). Wait, but then charge: left side: \(MnO_4^-\) is -1, \(H_2O\) is 0, total -1. Right side: \(MnO_2\) is 0, \(4OH^-\) is -4, total -4. So we need to balance charge. So we need to add electrons. But the question is just about the number of \(H_2O\) molecules, so atom balance (H and O) gives x=2? Wait, no, maybe I messed up.

Wait, another way: let's count the number of O atoms. On the product side, \(MnO_2\) has 2 O, \(4OH^-\) has 4 O, so total 6 O. On the reactant side, \(MnO_4^-\) has 4 O, so we need 2 more O, which comes from \(H_2O\) (each \(H_2O\) has 1 O). Wait, no, each \(H_2O\) has 1 O? No, \(H_2O\) has 1 O atom per molecule. Wait, \(H_2O\) has 1 O, so to get 2 more O, we need 2 \(H_2O\) (2 O). Then H: 2 \(H_2O\) has 4 H, which matches 4 \(OH^-\) (4 H). So yes, x=2? Wait, but the answer is 3? Wait, no, let's check the reaction again.

Wait, maybe the reaction is in basic medium, and we missed a step. Wait, let's do redox balancing properly.

1. Write half-reactions:

Reduction: \(MnO_4^- ightarrow MnO_2\) (in basic medium)

Oxidation: (assuming no oxidation, but maybe it's a reduction reaction)

Balance Mn: 1 \(MnO_4^-\) → 1 \(MnO_2\)

Balance O: add \(H_2O\) to right: \(MnO_4^- ightarrow MnO_2 + 2H_2O\) (now O: 4 on left, 2 + 4=6? No, \(MnO_2\) has 2 O, 2 \(H_2O\) has 2 O, total 4 O. Wait, no, \(MnO_4^-\) has 4 O, so to balance O, add 2 \(H_2O\) to right: \(MnO_4^- ightarrow MnO_2 + 2H_2O\) (O: 4 left, 2 + 2=4 right? No, \(MnO_2\) has 2 O, 2 \(H_2O\) has 2 O, total 4 O. Correct.

Now balance H: add \(OH^-\) to left and \(H_2O\) to right? Wait, basic medium. After balancing O, we have \(MnO_4^- ightarrow MnO_2 + 2H_2O\). Now balance H: left side has 0 H, right side has 4 H (from 2 \(H_2O\)). So add 4 \(OH^-\) to left and 2 \(H_2O\) to right? Wait, no, the formula is: in basic medium, after balancing O with \(H_2O\), balance H by adding \(H_2O\) to the side with less H and \(OH^-\) to the other side. Wait, maybe I confused.

Wait, the given product has 4 \(OH^-\), so let's use that. The product is \(MnO_2 + 4OH^-\). So let's balance H: 4 \(OH^-\) has 4 H, so we need 2 \(H_2O\) (4 H) on reactant. Then O: 2 \(H_2O\) (2 O) + \(MnO_4^-\) (4 O) = 6 O. Product: \(MnO_2\) (2 O) + 4 \(OH^-\) (4 O) = 6 O. Perfect. So the number of \(H_2O\) is 2? Wait, but when I check charge, reactant: \(MnO_4^-\) (-1) + 2 \(H_2O\) (0) = -1. Product: \(MnO_2\) (0) + 4 \(OH^-\) (-4) = -4. So we need to add 3 electrons to left: \(3e^- + MnO_4^- + 2H_2O ightarrow MnO_2 + 4OH^-\). Now charge: left: -1 + 0 + (-3) = -4. Right: 0 + (-4) = -4. Perfect. So the number of \(H_2O\) is 2? Wait, but the answer is 3? Wait, no, maybe I made a mistake.

Wait, let's count the O atoms again. Product: \(MnO_2\) (2 O) + 4 \(OH^-\) (4 O) = 6 O. Reactant: \(MnO_4^-\) (4 O) + x \(H_2O\) (x O). So 4 + x = 6 ⇒ x=2. H atoms: product has 4 H (from 4 \(OH^-\)), reactant has 2x H (from x \(H_2O\)). So 2x=4 ⇒ x=2. Yes, so x=2. Wait, but the initial thought was wrong about charge, but after adding electrons, it balances. So the number of \(H_2O\) molecules is 2? Wait, no, wait the problem is asking for the coefficient of \(H_2O\) on the reactant side. Let's check the reaction again:

\(2H_2O + MnO_4^- ightarrow MnO_2 + 4OH^-\)

Check O: 2 (from \(H_2O\)) + 4 (from \(MnO_4^-\)) = 6. Product: 2 (from \(MnO_2\)) + 4 (from \(OH^-\)) = 6. Good.

Check H: 4 (from 2 \(H_2O\)) = 4 (from 4 \(OH^-\)). Good.

Check charge: Left: \(MnO_4^-\) (-1) + 2 \(H_2O\) (0) = -1. Right: \(MnO_2\) (0) + 4 \(OH^-\) (-4) = -4. Wait, that's not balanced. Oh! I forgot to balance charge. So we need to add electrons. So \(3e^- + MnO_4^- + 2H_2O ightarrow MnO_2 + 4OH^-\). Now charge: Left: -1 + 0 + (-3) = -4. Right: 0 + (-4) = -4. Perfect. So the coefficient of \(H_2O\) is 2? Wait, but maybe the question is just about atom balance (H and O) and not charge, so the number of \(H_2O\) is 2? Wait, no, the answer is 3? Wait, I'm confused.

Wait, let's do it again. Let's list the number of each atom:

- Mn: 1 on left, 1 on right. Balanced.

- O: Left: \(MnO_4^-\) (4) + \(xH_2O\) (x) = 4 + x. Right: \(MnO_2\) (2) + \(4OH^-\) (4) = 6. So 4 + x = 6 ⇒ x=2.

- H: Left: \(xH_2O\) (2x). Right: \(4OH^-\) (4). So 2x = 4 ⇒ x=2.

Yes, so x=2. So the number of \(H_2O\) molecules is 2? Wait, but when I check the reaction with x=2, it works for H and O. The charge is balanced by electrons, but the question is about the number of \(H_2O\) molecules, so it's 2? Wait, no, maybe I made a mistake. Wait, the correct answer is 3? Wait, no, let's check with x=3.

If x=3:

O: 3 + 4 = 7. Product: 2 + 4 = 6. Not balanced.

H: 6 (from 3 \(H_2O\)) vs 4 (from 4 \(OH^-\)). Not balanced.

So x=2 is correct.

Answer:

2