Question

now lets look at the reduction half-reaction. how many water molecules should be added to the product side? $\ce{mno^{-}_{4} \
ightarrow mn^{2+} + ?h_{2}o}$

Explanation:

Step1: Balance oxygen atoms

In the reactant side, we have \( \text{MnO}_4^- \) which contains 4 oxygen atoms. On the product side, the oxygen atoms will be in \( \text{H}_2\text{O} \) molecules. Let the number of \( \text{H}_2\text{O} \) molecules be \( x \). So, the number of oxygen atoms from \( \text{H}_2\text{O} \) is \( x \) (since each \( \text{H}_2\text{O} \) has 1 O atom). We need to balance the O atoms, so \( x = 4 \)? Wait, no, wait. Wait, the reactant is \( \text{MnO}_4^- \) (1 Mn, 4 O), product side initially has \( \text{Mn}^{2+} \) (no O) and \( \text{H}_2\text{O} \). So to balance O, we need 4 O on product side, so number of \( \text{H}_2\text{O} \) molecules should be 4? Wait, no, let's do it properly.

Wait, the half - reaction is in acidic or basic medium? Wait, usually, for balancing redox half - reactions, in acidic medium, we balance O by adding \( \text{H}_2\text{O} \) and H by adding \( \text{H}^+ \). Let's assume acidic medium here.

The unbalanced half - reaction: \( \text{MnO}_4^-
ightarrow \text{Mn}^{2+} \)

Step 1: Balance Mn. Mn is already balanced (1 Mn on each side).

Step 2: Balance O. There are 4 O in \( \text{MnO}_4^- \), so we add 4 \( \text{H}_2\text{O} \) on the product side to balance O (since each \( \text{H}_2\text{O} \) has 1 O). So the reaction becomes \( \text{MnO}_4^-
ightarrow \text{Mn}^{2+}+ 4\text{H}_2\text{O} \)

Wait, let's check:

Reactant side: Mn: 1, O: 4

Product side after adding 4 \( \text{H}_2\text{O} \): Mn: 1, O: 4 (from 4 \( \text{H}_2\text{O} \)), H: 8 (from 4 \( \text{H}_2\text{O} \))

Then we can balance H by adding \( \text{H}^+ \) on the reactant side (for acidic medium). But the question is only about the number of \( \text{H}_2\text{O} \) molecules on the product side to balance O. Since there are 4 O in \( \text{MnO}_4^- \), we need 4 \( \text{H}_2\text{O} \) molecules (each with 1 O) to balance the O atoms.

Wait, but let's confirm. The formula is \( \text{MnO}_4^-
ightarrow \text{Mn}^{2+}+ x\text{H}_2\text{O} \)

Number of O atoms on left: 4

Number of O atoms on right: \( x \) (from \( x\text{H}_2\text{O} \))

So, to balance O, \( x = 4 \)

Step2: Verify

After adding 4 \( \text{H}_2\text{O} \) on the product side, the O atoms are balanced (4 O on left in \( \text{MnO}_4^- \) and 4 O on right in 4 \( \text{H}_2\text{O} \)).

Answer:

4