Balance this chemical equation.
Cr2O7 2-(aq) + Cu(s)---> Cr3+(aq) +Cu2+aq)
I don't understand how to do this problem. Where do i begin? thank you.
To balance a chemical equation, we want to make sure that the number of atoms on both sides of the equation is the same. We start by counting the number of atoms on each side of the equation.
On the left side of the equation (reactants):
Cr: 2
O: 7
H: 0 (not included in the equation)
Cu: 1
On the right side of the equation (products):
Cr: 1
O: 0
H: 0 (not included in the equation)
Cu: 2
Since the number of atoms of Cr and Cu are different on both sides, the equation is not balanced. We need to adjust the coefficients in front of the reactants and products to equalize the number of atoms on both sides.
Let's start by balancing Cr, which has 2 on the left and 1 on the right. We can balance Cr by placing a coefficient of 2 in front of Cr3+:
Cr2O7 2-(aq) + Cu(s) ---> 2Cr3+(aq) + Cu2+(aq)
Now, let's balance Cu, which has 1 on the left and 2 on the right. We can balance Cu by placing a coefficient of 2 in front of Cu(s):
Cr2O7 2-(aq) + 2Cu(s) ---> 2Cr3+(aq) + Cu2+(aq)
Now, let's balance the charges to make the equation electrically neutral. On the left side, the overall charge is -2, and on the right side, the overall charge is +2. We can balance the charges by adding two electrons (e-) to the left side:
Cr2O7 2-(aq) + 2Cu(s) + 14H+(aq) + 6e- ---> 2Cr3+(aq) + Cu2+(aq) + 7H2O(l)
Finally, the balanced chemical equation is:
Cr2O7 2-(aq) + 2Cu(s) + 14H+(aq) + 6e- ---> 2Cr3+(aq) + Cu2+(aq) + 7H2O(l)
Remember to always check that the equation is balanced by counting the number of atoms and charges on both sides.
To balance a chemical equation, you need to ensure that the number of atoms of each element is the same on both sides of the equation. Here's how to approach balancing this equation:
1. Start by counting the number of atoms of each element on both sides of the equation.
On the left side:
Cr: 2
O: 7
Cu: 1
On the right side:
Cr: 1
O: 0
Cu: 2
2. Begin by balancing the atoms of elements that appear in only one compound on each side. In this case, let's start with copper (Cu).
On the left side, there is one Cu atom, and on the right side, there are two Cu atoms. To balance it, place a coefficient of 2 in front of Cu on the left side:
Cr2O7 2-(aq) + 2Cu(s) ---> Cr3+(aq) + Cu2+(aq)
Now the number of Cu atoms is balanced with 2 on each side.
3. Next, balance the Cr atoms. On the left side, there are two Cr atoms, and on the right side, there is only one Cr atom. To balance it, place a coefficient of 2 in front of Cr3+ on the right side:
Cr2O7 2-(aq) + 2Cu(s) ---> 2Cr3+(aq) + Cu2+(aq)
Now the number of Cr atoms is balanced with 2 on each side.
4. Finally, balance the oxygen (O) atoms. On the left side, there are 7 O atoms, and on the right side, there are no O atoms. To balance it, add 7 H2O molecules to the right side:
Cr2O7 2-(aq) + 2Cu(s) ---> 2Cr3+(aq) + Cu2+(aq) + 7H2O(l)
Now the number of O atoms is balanced with 7 on each side.
And that's it! The final balanced chemical equation is:
Cr2O7 2-(aq) + 2Cu(s) ---> 2Cr3+(aq) + Cu2+(aq) + 7H2O(l)
Where do you start? By assigning oxidation numbers to each of the elements to determine which have changed. Here is a hint to get you started. Cr changes from +6 for EACH Cr on the left to +3 on the right. Cu changes from zero on the left to +2 on the right.
You need to learn to do these redox equations yourself. Here is a site that tells you almost anything you want to know about redox equation.
http://www.chemteam.info/Redox/Redox.html
If you have specific questions as you go through I'll be more than happy to help you through.