Posted by Carl on Tuesday, April 27, 2010 at 7:58pm.
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.
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