I cannot balance this equation and I am really struggling.
Cr2O7^2-(aq)+ Cu^+(s)-->Cr^3+(aq)+Cu^2+(aq)
How do I get the charges to balance?
1. Place a 2 for Cr^+3. This is a temporary 2 and it may be changed later but you MUST start with the same number before counting electrons lost or gained or you will NEVER get it to balance.
2. Now you count total charge on Cr in Cr2O7^-2 and that is +12. For Cr on the right we have 2Cr^+3 or +6 total. Therefore, Cr gained 6 electrons to go from +12 to +6 for the two Cr atoms.
3. Cu is just one on each side. That is +1 on the left (but I DON'T think it is a solid) [or if it is then it is zero]. Assuming +1 on left and +2 on right is a loss of 1 electron.
f. So you multiply the Cu^+1/Cu^+2 by 6 and the Cr2O7^-2/2Cr^+3 by 1 (which also means the temporary 2 for Cr^+ is now a permanent 2).
4. Now you can add H^+ and H2O to balance the oxygen atoms and charges.
To balance the charges in this chemical equation, you need to ensure that the total positive charge on each side of the equation is equal to the total negative charge. Here's how you can go about it:
1. Identify the oxidation states of the elements:
The oxidation state of Cr in Cr2O7^2- is +6, while Cu has an oxidation state of +1 in Cu+ and +2 in Cu2+.
2. Split the Cr2O7^2- ion into its constituent atoms:
Cr2O7^2- can be split into two Cr^3+ ions and seven O^2- ions.
The equation now becomes:
2Cr^3+(aq) + Cu^+(s) --> 2Cr^3+(aq) + Cu^2+(aq) + 7O^2-(aq)
3. Check the charges:
On the left-hand side, we have a total charge of +4 (2x +3 for Cr^3+ and +1 for Cu^+).
On the right-hand side, we have a total charge of +4 (2x +3 for Cr^3+ and 2x +2 for Cu^2+).
The charges on both sides are balanced.
Therefore, the balanced equation is:
2Cr2O7^2-(aq) + 3Cu^+(s) --> 4Cr^3+(aq) + 3Cu^2+(aq) + 7O^2-(aq)