Guys i really need help in these three problems. If someone could explain how to do this i would appreciate it!
complete and balance the following equations
1. SO3^2- + Cu^2+ --> SO4^2- + Cu^+
2. Cr +Cu^2+ ---> CrO4^2- + Cu^+
3. MnO2 + Ag^+ ---> MnO4^- + Ag
Here is a web site that tells you how to do this. To balanced these you need
1. how to determine the oxidation state of elements.
2. how to determine the loss and gain of electrons.
3. the system for balancing half equations.
4. how to add the oxidation half to the reduction half.
http://www.chemteam.info/Redox/Redox.html
Sure, I can help explain how to complete and balance these chemical equations.
1. SO3^2- + Cu^2+ --> SO4^2- + Cu^+
To balance this equation, you need to make sure that the number of atoms of each element is the same on both sides of the equation.
We have:
1 sulfur (S) atom on the left side, 1 on the right side.
3 oxygen (O) atoms on the left side, 4 on the right side.
1 copper (Cu) atom on the left side, 1 on the right side.
To balance the sulfur (S) atoms, we need to put a coefficient of 1 in front of the SO4^2- on the right side. This gives us:
SO3^2- + Cu^2+ --> 1 SO4^2- + Cu^+
To balance the oxygen (O) atoms, we need to put a coefficient of 2 in front of the SO3^2- on the left side. Similarly, we need to put a coefficient of 3 in front of the Cu^+ on the right side. This gives us:
2 SO3^2- + Cu^2+ --> 1 SO4^2- + 3 Cu^+
Now the equation is balanced with respect to sulfur (S), oxygen (O), and copper (Cu) atoms.
2. Cr + Cu^2+ ---> CrO4^2- + Cu^+
To balance this equation, we need to follow the same steps as before.
We have:
1 chromium (Cr) atom on the left side, 1 on the right side.
1 copper (Cu) atom on the left side, 1 on the right side.
For oxygen (O), we have 4 on the right side but none on the left side.
To balance the chromium (Cr) atoms, we need to put a coefficient of 2 in front of the CrO4^2- on the right side.
This gives us:
Cr + Cu^2+ --> 2 CrO4^2- + Cu^+
Now the equation is balanced with respect to chromium (Cr) and copper (Cu) atoms.
3. MnO2 + Ag^+ ---> MnO4^- + Ag
For this equation, we have:
1 manganese (Mn) atom on the left side, 1 on the right side.
2 oxygen (O) atoms on the left side, 4 on the right side.
1 silver (Ag) atom on the left side, 1 on the right side.
To balance the oxygen (O) atoms, we need to put a coefficient of 2 in front of the Ag^+ on the left side. This gives us:
MnO2 + 2 Ag^+ ---> MnO4^- + Ag
Now the equation is balanced with respect to manganese (Mn), oxygen (O), and silver (Ag) atoms.
To summarize, the balanced equations are:
1. 2 SO3^2- + Cu^2+ --> SO4^2- + 3 Cu^+
2. Cr + Cu^2+ --> 2 CrO4^2- + Cu^+
3. MnO2 + 2 Ag^+ ---> MnO4^- + Ag