Question

the two half - reactions are given below.
$7e^- + 4h_2o + reo_4^- \
ightarrow re + 8oh^-$
$4oh^- + io^- \
ightarrow io_3^- + 2h_2o + 4e^-$
what is the coefficient for $h_2o$ in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in both half - reactions

The first half - reaction: \(7e^-+4H_2O + ReO_4^- ightarrow Re + 8OH^-\) (gain of 7 electrons)
The second half - reaction: \(4OH^-+IO^- ightarrow IO_3^-+2H_2O + 4e^-\) (loss of 4 electrons)

To balance the number of electrons, we multiply the first half - reaction by 4 and the second half - reaction by 7.

First half - reaction (after multiplying by 4):
\(4\times(7e^-+4H_2O + ReO_4^- ightarrow Re + 8OH^-)\)
\(28e^-+16H_2O + 4ReO_4^- ightarrow 4Re + 32OH^-\)

Second half - reaction (after multiplying by 7):
\(7\times(4OH^-+IO^- ightarrow IO_3^-+2H_2O + 4e^-)\)
\(28OH^-+7IO^- ightarrow 7IO_3^-+14H_2O + 28e^-\)

Step2: Add the two half - reactions together

Now, add the two modified half - reactions:

\(28e^-+16H_2O + 4ReO_4^-+28OH^-+7IO^- ightarrow 4Re + 32OH^-+7IO_3^-+14H_2O + 28e^-\)

Cancel out the electrons (\(28e^-\) on both sides). Also, cancel out the \(OH^-\) and \(H_2O\) that can be simplified.

For \(OH^-\): \(28OH^-\) on the left and \(32OH^-\) on the right. After cancellation, we have \(32 - 28=4OH^-\) on the right.

For \(H_2O\): \(16H_2O\) on the left and \(14H_2O\) on the right. After cancellation, we have \(16 - 14 = 2H_2O\) on the left.

The overall reaction after cancellation:

\(16H_2O-14H_2O+4ReO_4^-+7IO^-+28OH^- ightarrow 4Re + 7IO_3^-+32OH^-\)

\(2H_2O+4ReO_4^-+7IO^- ightarrow 4Re + 7IO_3^-+4OH^-\) Wait, no, let's do the addition again carefully.

Wait, when we add the two reactions:

Left side \(H_2O\): 16 \(H_2O\)

Right side \(H_2O\): 14 \(H_2O\)

So net \(H_2O\) on left: \(16 - 14=2\)? Wait, no, let's re - check the addition of \(H_2O\):

Left \(H_2O\): 16

Right \(H_2O\): 14

So when we add the two reactions, the \(H_2O\) term is \(16H_2O\) (left) and \(14H_2O\) (right). So the net \(H_2O\) in the final equation:

\(16H_2O-14H_2O = 2H_2O\) on the left? Wait, no, let's do the addition step by step.

Wait, the correct way is:

Left side: \(16H_2O\) (from first half - reaction)

Right side: \(14H_2O\) (from second half - reaction)

So when we combine the two reactions, the \(H_2O\) in the overall reaction:

\(16H_2O\) (left) and \(14H_2O\) (right). So the coefficient of \(H_2O\) is \(16 - 14=2\)? Wait, no, maybe I made a mistake in multiplication.

Wait, first half - reaction: electrons gained = 7, second: electrons lost = 4. LCM of 7 and 4 is 28. So multiply first by 4, second by 7.

First half - reaction (multiplied by 4):

\(28e^-+16H_2O + 4ReO_4^- ightarrow 4Re + 32OH^-\)

Second half - reaction (multiplied by 7):

\(28OH^-+7IO^- ightarrow 7IO_3^-+14H_2O + 28e^-\)

Now, add them:

\(28e^-+16H_2O + 4ReO_4^-+28OH^-+7IO^- ightarrow 4Re + 32OH^-+7IO_3^-+14H_2O + 28e^-\)

Now, subtract the right - side \(H_2O\) from the left - side \(H_2O\): \(16H_2O-14H_2O = 2H_2O\) (left)

Subtract the left - side \(OH^-\) from the right - side \(OH^-\): \(32OH^--28OH^- = 4OH^-\) (right)

So the overall equation is:

\(2H_2O+4ReO_4^-+7IO^- ightarrow 4Re + 7IO_3^-+4OH^-\) Wait, no, that can't be right. Wait, maybe I messed up the \(H_2O\) calculation.

Wait, let's do the addition again:

Left side: \(16H_2O\) (from first half - reaction)

Right side: \(14H_2O\) (from second half - reaction)

So the net \(H_2O\) is \(16 - 14 = 2\)? Wait, no, when adding, we have \(16H_2O\) on the left and \(14H_2O\) on the right. So in the final equation, the coefficient of \(H_2O\) is \(16-14 = 2\)? Wait, no, maybe I made a mistake in the multiplication of the half - reactions.

Wait, let's re - examine the half - reactions.

First half - reaction: \(7e^-+4H_2O+ReO_4^- ightarrow Re + 8OH^-\) (reduction, gain 7e⁻)

Second half - reaction: \(4OH^-+IO^- ightarrow IO_3^-+2H_2O + 4e^-\) (oxidation, lose 4e⁻)

To balance electrons, multiply reduction by 4 and oxidation by 7.

Reduction (×4): \(28e^-+16H_2O + 4ReO_4^- ightarrow 4Re + 32OH^-\)

Oxidation (×7): \(28OH^-+7IO^- ightarrow 7IO_3^-+14H_2O + 28e^-\)

Now, add the two equations:

\(28e^-+16H_2O + 4ReO_4^-+28OH^-+7IO^- ightarrow 4Re + 32OH^-+7IO_3^-+14H_2O + 28e^-\)

Now, cancel \(28e^-\) from both sides.

For \(OH^-\): 28 on left, 32 on right. So 32 - 28 = 4 \(OH^-\) on right.

For \(H_2O\): 16 on left, 14 on right. So 16 - 14 = 2 \(H_2O\) on left.

So the balanced equation is:

\(2H_2O+4ReO_4^-+7IO^- ightarrow 4Re + 7IO_3^-+4OH^-\)

Wait, but that seems odd. Wait, maybe I made a mistake in the direction of \(H_2O\). Let's check the \(H\) and \(O\) atoms.

In the first half - reaction (×4): 16 \(H_2O\) (32 H, 16 O) and 32 \(OH^-\) (32 H, 32 O) on the right.

In the second half - reaction (×7): 28 \(OH^-\) (28 H, 28 O) on the left, 14 \(H_2O\) (28 H, 14 O) on the right.

Total H on left: 32 (from \(H_2O\))+28 (from \(OH^-\)) = 60 H

Total H on right: 32 (from \(OH^-\))+28 (from \(H_2O\)) = 60 H

Total O on left: 16 (from \(H_2O\))+4×4 (from \(ReO_4^-\))+28 (from \(OH^-\))+7 (from \(IO^-\)) = 16 + 16+28 + 7=67 O

Total O on right: 32 (from \(OH^-\))+7×3 (from \(IO_3^-\))+14 (from \(H_2O\)) = 32+21 + 14=67 O

Now, when we add the two equations, the \(H_2O\) on the left is 16, on the right is 14. So the net \(H_2O\) is 16 - 14 = 2. So the coefficient of \(H_2O\) in the balanced net ionic equation is 2? Wait, no, maybe I messed up. Wait, let's do the addition again.

Wait, the two half - reactions after multiplying:

Reaction 1: \(28e^-+16H_2O + 4ReO_4^- ightarrow 4Re + 32OH^-\)

Reaction 2: \(28OH^-+7IO^- ightarrow 7IO_3^-+14H_2O + 28e^-\)

Now, add them:

Left side: \(28e^-+16H_2O + 4ReO_4^-+28OH^-+7IO^-\)

Right side: \(4Re + 32OH^-+7IO_3^-+14H_2O + 28e^-\)

Now, subtract the right - side terms from the left - side terms for like species:

For \(H_2O\): \(16H_2O-14H_2O = 2H_2O\) (left)

For \(OH^-\): \(28OH^--32OH^-=- 4OH^-\) (which means 4 \(OH^-\) on the right)

For electrons: \(28e^--28e^- = 0\)

So the equation becomes:

\(2H_2O+4ReO_4^-+7IO^- ightarrow 4Re + 7IO_3^-+4OH^-\)

Wait, but let's check the number of O atoms in \(H_2O\):

Left side \(H_2O\): 2×1 = 2 O

Right side \(OH^-\): 4×1 = 4 O; \(IO_3^-\): 7×3 = 21 O; \(ReO_4^-\): 4×4 = 16 O; \(IO^-\):7×1 = 7 O

Left side O: 2 (from \(H_2O\))+16 (from \(ReO_4^-\))+7 (from \(IO^-\)) = 25 O

Right side O: 4 (from \(OH^-\))+21 (from \(IO_3^-\)) = 25 O. It balances.

Number of H atoms:

Left side \(H_2O\): 2×2 = 4 H; \(OH^-\): 0 (since we canceled them out? Wait, no, in the left side after addition, we had 28 \(OH^-\) and in the right side 32 \(OH^-\). Wait, I think I made a mistake in the \(OH^-\) calculation.

Wait, left side \(OH^-\): 28 (from reaction 2)

Right side \(OH^-\): 32 (from reaction 1)

So when we add the two reactions, the \(OH^-\) on the left is 28, on the right is 32. So the net \(OH^-\) is 32 - 28 = 4 on the right.

Left side \(H_2O\): 16 (from reaction 1)

Right side \(H_2O\):14 (from reaction 2)

Net \(H_2O\):16 - 14 = 2 on the left.

Yes, that seems correct. So the coefficient of \(H_2O\) is 2. Wait, but let's check with another approach.

Alternative approach:

After multiplying the first half - reaction by 4 and the second by 7, we have:

Reaction 1: \(28e^-+16H_2O + 4ReO_4^- ightarrow 4Re + 32OH^-\)

Reaction 2: \(28OH^-+7IO^- ightarrow 7IO_3^-+14H_2O + 28e^-\)

Now, add the two reactions:

\(16H_2O+4ReO_4^-+28OH^-+7IO^- ightarrow 4Re + 32OH^-+7IO_3^-+14H_2O\)

Now, subtract \(14H_2O\) from both sides: \(16H_2O - 14H_2O=2H_2O\), so \(2H_2O+4ReO_4^-+28OH^-+7IO^- ightarrow 4Re + 32OH^-+7IO_3^-\)

Subtract \(28OH^-\) from both sides: \(32OH^--28OH^- = 4OH^-\), so \(2H_2O+4ReO_4^-+7IO^- ightarrow 4Re + 4OH^-+7IO_3^-\)

Yes, so the coefficient of \(H_2O\) is 2.

Answer:

2