1. What is the balanced form of the following equation? Br2 + S2O32– + H2O ? Br1– + SO42– + H+

Question

Please give me step by step and final answer

Answer 1

Mia, to balance equations with charges it's important to balance the atoms first. Let's start by listing the number of each atom on both sides.

 

Br₂ + S₂O₃ ^2- + H₂O –> Br ^- + SO₄ + H⁺ is the equation, right?

So, if we were to count the number of atoms on both sides, it would be: 2 Br, 2S, 3O, 2H, 1O | 1 Br, 1S, 4O, 1H. Because O appears with S and H on the reactant side of the equation, you would combine the count. This means that instead of 3O and 1O its 4O (3+1=4 so 4 oxygen atoms). So the count is: 2 Br, 2S, 4O, 2H | 1 Br, 1S, 4O, 1H. 

 

Now it helps to look at the atoms that have the same number on both sides of the equation so you can see which atoms you don't have to worry about. The O count is the same on both sides, but since it's associated with S and H on the reactant side of the equation, you do have to worry about it on the product side. All right so far?

 

It's also easier to leave the solitary atoms alone until the end, so here we'll fix the count for Br towards the end since its alone on both sides of the equation. For right now, let's look at S. As part of the reactant, there are 2 of them, but only 1 as part of the product. This means that we must have another S atom on the product side, so the equation would be:

 

Br2 + S2O3 2- + H2O –> Br- + 2SO4 + H+    with the count as 2 Br, 2S, 4O, 2H | 1 Br, 2S, 4O, 1H

However, this also affects O on the product side, making the count be 8O since 2•4=8.

The count is thus 2 Br, 2S, 4O, 2H | 1 Br, 2S, 8O, 1H.

 

At this point, the count is incorrect for O, Br, and H. To fix the count for O, you need to increase the number of O atoms without adjusting S. To do that, you would only increase the number of water molecules instead of the S2O3 molecule. You need 8 O on the reactant side and so far you only have 4, so that means you must have 5H2O molecules (it would give you 5O in addition to the 3 from S2O3). Now the O count is set and the equation look like this:

Br2 + S2O3 2- + 5H2O –> Br- + 2SO4 + H+   with the count   2 Br, 2S, 8O, 5H | 1 Br, 2S, 8O, 1H.

 

You now need to adjust the H and Br count. On the product side, you have 1 H as well as 1 Br when you need 5H and 2Br. The adjustments are made on the product side and because both atoms are by themselves this makes it easy! Just have 10H and 2Br on the product side so the atoms are balanced (remember that its H2 on the reactant side)! The equation now looks like this:

 

Br2 + S2O3 2- + 5H2O –> 2Br- + 2SO4 + 10H+  with the count 2 Br, 2S, 8O, 10H | 2 Br, 2S, 8O, 10H. As you can see, the count shows you that both sides have the same amount of each type of atom. Now we need to look at the charges.

 

To figure out the amount of charge on each side, just look at the exponential number (small number above last atom of molecule) and add them together. On the reactant side, there is only a charge of -2 because of one S2O3 with a charge of -2 (1• -2= -2); Br2 and 5H2O are inert, charge-wise.

 

However, on the product side there is a total charge of 4. This is because 2•-1 + 2•-2 + 10•1= -2 + -4 +10= -6 +10= 4. To make the charge balanced, you need to adjust the side with the larger number to make it equal the smaller one. This is because of the electrons which have a negative charge, thus bring the number down. So how do you get 4 to become -2? You would subtract 6, so this means that there are 6 electrons on the product side.

 

All together, the equation looks like this:

 

Br₂ + S₂O₃^2- + 5H₂O –> 2Br^- + 2SO₄ + 10H⁺ + 6e

 

I hope this helps!

 

 

 

Answer 2

Br₂+S₂O₃^2-+5H₂0  -->  2Br^-1+2SO₄^-2+10H⁺