I get that you're supposed to balance; however, I was confused on how to do the reduction and oxidation electron equations (the simplist way). Looked at too many examples online but nothing was clear.
1) Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
2.) KMnO4 + H2SO4 + FeSO4 → MnSO4 + Fe2(SO4)3 + K2SO4 + H2O
1 is a little tricky. They way you do it is to balance the redox part, then the tricky part.
Cu is 0 on the left and +2 on the right. Draw a line from one to the other and show loss of 2e.
Then you look at N. It is +5 on the left and it shows +5 on the right (with the NO3^-) and +4 on the right from NO2. In other words it changed but it didn't change. :-). So ignore, for now, the part that didn't change (since that has nothing to do with the redox part) and focus on the NO2. Draw a line from the HNO3 to NO2 and show gain of 1e. Now make the loss = gain SO multiply the HNO3/NO2 by 2 and you get this.
Cu + 2HNO3 ==> Cu(NO3)2 + 2NO2 + H2O and that balances the redox part. Now, the tricky part, is that we know some of the NO3^- did not change so we take care of that now. We look and see Cu(NO3)2 on the right so there are two NO3^- that didn't change to NO2 which means we must add another 2HNO3 to get those 2NO3^- to go with the Cu^2+. We simply add those two manually and it looks like this.
Cu + 4HNO3 ==> Cu(NO3)2 + 2NO2 + H2O. Then we check it, and see that changes the H2O and we balance that manually, too.
Cu + 4HNO3 ==> Cu(NO3)2 + 2NO2 + 2H2O
All done.
The place students mess up on the #2 problem is they start off wrong which means they can spend hours thinking they are following the rules and they never get it balanced. Here is the #1 move. Place a 2 as the coefficient for FeSO4. Then balance normally. You can't compare apples and oranges. Since there are 2 Fe on the right from the Fe2(SO4)3 and only one from FeSO4 on the left. you MUST compare the same number of atoms and the total electron change. I'll eave that for you. By the way, the chem team gives a good set of instructions. Here is their web site. Click on the part you want.
http://www.chemteam.info/Redox/Redox.html
Really appreciate it! Thanks for the break down.
Balancing reduction-oxidation (redox) equations can be challenging, but there are some steps you can follow to simplify the process. Here's a step-by-step guide to balancing the given equations:
1) Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Step 1: Identify the elements undergoing oxidation and reduction. In this equation, Cu is being oxidized from 0 to +2, and N in HNO3 is being reduced from +5 to +2.
Step 2: Start by balancing the atoms other than oxygen and hydrogen. Here, we have one Cu atom on the left side, so we place a coefficient of 1 in front of Cu on the right side.
Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Step 3: Balance the oxygen atoms by adding water molecules (H2O) to the side that needs more oxygen. In this case, we need two oxygen atoms on the left side, so we add a coefficient of 2 in front of H2O.
Cu + HNO3 → Cu(NO3)2 + NO2 + 2H2O
Step 4: Balance the hydrogen atoms by adding H+ ions. In this equation, we need eight hydrogen atoms on the left side. Therefore, we add eight H+ ions to the right side.
Cu + HNO3 → Cu(NO3)2 + NO2 + 2H2O + 8H+
Step 5: Balance the charges by adding electrons. On the left side of the equation, Copper (Cu) goes from an oxidation state of 0 to +2, so two electrons (2e-) are added to the left side.
Cu + 2e- + HNO3 → Cu(NO3)2 + NO2 + 2H2O + 8H+
Step 6: Balance the electrons on the right side by multiplying some or all of the species by a factor to make the number of electrons equal on both sides. In this case, multiplying the NO3- ion by two will balance the electrons.
Cu + 2e- + HNO3 → Cu(NO3)2 + NO2 + 2H2O + 8H+
Step 7: Finally, make sure that the number of atoms and charges are balanced on each side. In this equation, we have 2 Cu, 2 N, 6 O, 8 H, and 2 electrons on both sides, so the equation is balanced.
2) KMnO4 + H2SO4 + FeSO4 → MnSO4 + Fe2(SO4)3 + K2SO4 + H2O
Step 1: Identify the elements undergoing oxidation and reduction. In this equation, Mn is being reduced from +7 to +2, and Fe in FeSO4 is being oxidized from +2 to +3.
Step 2: Start by balancing the atoms other than oxygen and hydrogen. We have one Mn atom on the right side, so we place a coefficient of 1 in front of MnSO4 on the left side.
KMnO4 + H2SO4 + FeSO4 → MnSO4 + Fe2(SO4)3 + K2SO4 + H2O
Step 3: Balance the oxygen atoms by adding water molecules (H2O) to the side that needs more oxygen. In this case, we need four oxygen atoms on the left side, so we add a coefficient of 4 in front of H2O.
KMnO4 + H2SO4 + FeSO4 → MnSO4 + Fe2(SO4)3 + K2SO4 + 4H2O
Step 4: Balance the hydrogen atoms by adding H+ ions. We need eight hydrogen atoms on the right side, so we add eight H+ ions to the left side.
KMnO4 + H2SO4 + FeSO4 + 8H+ → MnSO4 + Fe2(SO4)3 + K2SO4 + 4H2O
Step 5: Balance the charges by adding electrons. In this equation, Mn in MnO4- gains five electrons to go from an oxidation state of +7 to +2.
KMnO4 + H2SO4 + FeSO4 + 8H+ + 5e- → MnSO4 + Fe2(SO4)3 + K2SO4 + 4H2O
Step 6: Balance the electrons on the right side by multiplying some or all of the species by a factor to make the number of electrons equal on both sides. In this case, multiplying the FeSO4 by 5 and the MnSO4 by 2 will balance the electrons.
KMnO4 + H2SO4 + 5FeSO4 + 8H+ + 5e- → 2MnSO4 + Fe2(SO4)3 + K2SO4 + 4H2O
Step 7: Make sure that the number of atoms and charges are balanced on each side. In this case, we have one Mn, 10 Fe, 2 S, 36 O, 12 H, and 5 electrons on both sides, so the equation is balanced.
Remember, practice is key when balancing redox equations. Keep analyzing examples, and over time, you'll become more familiar with the patterns and techniques involved.