Question

the redox reaction given below occurs in basic solution.
br⁻ + mno₄⁻ → bro₃⁻ + mno₂
this first half - reaction needs two more oxygen atoms on the product side. how many h₂o molecules should be added to the product side?
mno₄⁻ → mno₂ + ?h₂o

Explanation:

Step1: Analyze oxygen atoms

In the half - reaction \(MnO_4^- ightarrow MnO_2 +?H_2O\), we first count the number of oxygen atoms on each side. On the left - hand side (reactant side), the number of oxygen atoms in \(MnO_4^-\) is 4. On the right - hand side, the number of oxygen atoms in \(MnO_2\) is 2. Let the number of \(H_2O\) molecules be \(x\). The number of oxygen atoms from \(H_2O\) is \(x\) (since each \(H_2O\) has 1 oxygen atom). So the total number of oxygen atoms on the right - hand side is \(2 + x\).
We want the number of oxygen atoms on both sides to be equal. So we set up the equation based on the conservation of oxygen atoms: \(4=2 + x\).

Step2: Solve for x

We solve the equation \(4 = 2+x\) for \(x\). Subtract 2 from both sides of the equation: \(x=4 - 2=2\). But wait, the problem states that "This first half - reaction needs two more oxygen atoms on the product side". Let's re - examine the half - reaction. The reactant is \(MnO_4^-\) (4 O atoms) and the product \(MnO_2\) has 2 O atoms. We need to add O atoms to the product side. Each \(H_2O\) molecule provides 1 O atom. To get 2 more O atoms on the product side (since \(4-2 = 2\) O atoms are needed on the product side to balance the O atoms from the reactant), we need to add 2 \(H_2O\) molecules? Wait, no. Wait, the reactant has 4 O, product \(MnO_2\) has 2 O. We need the product side to have 4 O (to balance with reactant's 4 O) from \(MnO_2\) and \(H_2O\). So the number of O atoms from \(H_2O\) should be \(4 - 2=2\). Since each \(H_2O\) has 1 O atom, the number of \(H_2O\) molecules is 2? Wait, no, let's do it step by step.

Wait, the half - reaction is \(MnO_4^- ightarrow MnO_2 +?H_2O\). Let's count O atoms:

Left: 4 O (in \(MnO_4^-\))

Right: 2 O (in \(MnO_2\))+\(x\) O (in \(x\) \(H_2O\) molecules, since each \(H_2O\) has 1 O).

For O - atom balance:

Number of O on left = Number of O on right

\(4=2 + x\)

Solving for \(x\): \(x = 4-2=2\). But the problem says "This first half - reaction needs two more oxygen atoms on the product side". The product side currently has 2 O (from \(MnO_2\)). To get two more O atoms (so that total O on product side is \(2 + 2=4\), which is equal to O on reactant side), we need to add 2 \(H_2O\) molecules (since each \(H_2O\) gives 1 O atom). Wait, but let's check again. Wait, the reactant is \(MnO_4^-\) (4 O), product \(MnO_2\) (2 O). We need to add O to the product side. Each \(H_2O\) has 1 O. So to get 2 more O atoms (because \(4-2 = 2\) O atoms are missing on the product side), we need 2 \(H_2O\) molecules. But wait, the problem says "This first half - reaction needs two more oxygen atoms on the product side". Let's see:

Reactant O: 4

Product O (before adding \(H_2O\)): 2

We need product O to be \(2 + 2=4\) (to match reactant O). So the number of O atoms to add is 2. Since each \(H_2O\) provides 1 O atom, the number of \(H_2O\) molecules is 2? Wait, no, wait a second. Wait, maybe I made a mistake. Let's think about the half - reaction balancing in basic solution. Wait, the standard way to balance O atoms in a half - reaction in basic solution:

For the half - reaction \(MnO_4^- ightarrow MnO_2\):

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

2. Balance O: Reactant has 4 O, product has 2 O. To balance O, we add \(H_2O\) to the side that needs O. Since the product side has less O, we add \(H_2O\) to the product side. The number of \(H_2O\) molecules added is equal to the number of O atoms needed. The number of O atoms needed on the product side is \(4 - 2 = 2\). Since each \(H_2O\) provides 1 O atom, we need to add 2 \(H_2O\) molecules? Wait, no, each \(H_2O\) has 1 O, so to get 2 O atoms, we need 2 \(H_2O\) molecules. Wait, but let's check the oxygen count:

After adding 2 \(H_2O\) to the product side:

Reactant O: 4 (from \(MnO_4^-\))

Product O: 2 (from \(MnO_2\))+2 (from 2 \(H_2O\)) = 4. So it is balanced.

Wait, but the problem says "This first half - reaction needs two more oxygen atoms on the product side". The product side initially has 2 O (from \(MnO_2\)). To get two more O atoms (so total 4 O, same as reactant), we need to add 2 \(H_2O\) molecules (since each \(H_2O\) gives 1 O atom).

Answer:

2