BrO3^-(aq)+Sn^2+(aq)+ ____ -->Br^-(aq)+Sn^4+(aq)+ _____
H20(l)and H^+(aq) are involved in the reaction.
What are the coefficients of the six species in the balanced equation above? Remember to include coefficients for H2O(l) and H+(aq) in the appropriate blanks.
This is an oxidation reduction equation and you need to learn how to do them. There is a systematic process to go through. I can help you get started.
Br goes from +5 on the left to -1 on the right. Sn goes from +2 on the left to +4 on the right. I am sure your text and or your notes will give the systematic approach. If not, I will try to find a set of good instructions on the web.
To balance the given equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's how to balance it step by step:
1. Start by balancing the atoms that appear in only one species on each side of the equation. In this case, we have Br and Sn on the right side, so we start with those.
BrO3^-(aq) + Sn^2+(aq) --> Br^-(aq) + Sn^4+(aq) + ...
2. Balance the bromine (Br) atoms by adding a coefficient of 3 in front of Br^-(aq):
BrO3^-(aq) + Sn^2+(aq) --> 3Br^-(aq) + Sn^4+(aq) + ...
3. Now balance the Sn (tin) atoms by adding a coefficient of 2 in front of Sn^2+(aq):
BrO3^-(aq) + 2Sn^2+(aq) --> 3Br^-(aq) + Sn^4+(aq) + ...
4. Next, balance the oxygen (O) atoms. We have 3 oxygens from BrO3^-(aq), so we need to add 3 waters (H2O):
BrO3^-(aq) + 2Sn^2+(aq) --> 3Br^-(aq) + Sn^4+(aq) + 3H2O(l)
5. Finally, balance the hydrogen (H) atoms. We have 6 hydrogens from the water (3 waters x 2H per water), so we need to add 6 hydrogen ions (H+):
BrO3^-(aq) + 2Sn^2+(aq) + 6H+(aq) --> 3Br^-(aq) + Sn^4+(aq) + 3H2O(l)
Therefore, the coefficients for the six species in the balanced equation are:
1, 2, 6, 3, 1, 3