Balance the following equation. (for a balanced eq. aA + bB → cC + dD, enter your answer as the integer abcd)
CrO42−(aq) + H2O(l) + HSnO2− → CrO2−(aq) + OH−(aq) + HSnO3−(aq)
3 answers
Tell me what you know to do and I can go from there. Do you have a system; there are two or three available and all of them work very well. Let me know the system too so I won't switch systems on you.
I know how to balance redox reactions but the charges are weird and I don't know where I'm going wrong or what to do
OK. Remember that compounds must add up with oxidation states to zero; ions must add up to the charge on the ion.
So Cr on the left is + 6 EACH and on the right is 3+.(O is -2 and there are two of them which makes -4 total so Cr must be +3 to leave a -1 charge on the CrO2^- ion). Sn on the left is 2+ and on the right is 4+. Let me know if you don't get this. I'm guessing here but years of teaching tells me that your trouble is that you don't START by placing a two as a coefficient for the CrO2^-; i.e., Cr2O7^2- ==> 2CrO2^- so that the TOTAL Cr on the left is 12+ and on the right is 6+ so you have a gain of 6 electrons for that half. Sn is loss of 2e for that half. So the multiplier is 1 for the Cr and 3 for the Sn. Keep in touch if you still have trouble.
So Cr on the left is + 6 EACH and on the right is 3+.(O is -2 and there are two of them which makes -4 total so Cr must be +3 to leave a -1 charge on the CrO2^- ion). Sn on the left is 2+ and on the right is 4+. Let me know if you don't get this. I'm guessing here but years of teaching tells me that your trouble is that you don't START by placing a two as a coefficient for the CrO2^-; i.e., Cr2O7^2- ==> 2CrO2^- so that the TOTAL Cr on the left is 12+ and on the right is 6+ so you have a gain of 6 electrons for that half. Sn is loss of 2e for that half. So the multiplier is 1 for the Cr and 3 for the Sn. Keep in touch if you still have trouble.
1 answer