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)
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.
To balance the given chemical equation, we need to ensure that both sides of the equation have an equal number of atoms of each element.
Let's start by counting the number of atoms for each element on both sides of the equation:
On the left side (reactants):
Cr: 1
O: 5
H: 3
Sn: 1
On the right side (products):
Cr: 1
O: 5
H: 3
Sn: 1
Since the number of atoms for each element is already balanced, we can skip adjusting the subscript numbers and move on to balancing the charges.
On the left side:
CrO4^(2-): -2
H2O: 0
HSnO2^-(aq): -1
On the right side:
CrO2^-(aq): -2
OH^-(aq): -1
HSnO3^-(aq): -1
To balance the charges, we can add H+ ions or H2O molecules to the appropriate side of the equation. Let's add H+ ions to the left side to balance the charge of HSnO2^-(aq):
CrO4^(2-)(aq) + H2O(l) + HSnO2^-(aq) + H+(aq) → CrO2^-(aq) + OH^-(aq) + HSnO3^-(aq)
Now let's count the number of atoms for each element again:
On the left side:
Cr: 1
O: 9
H: 4
Sn: 1
On the right side:
Cr: 1
O: 3
H: 2
Sn: 1
Now the number of atoms is not balanced, particularly for oxygen (O). To balance oxygen, we can add water (H2O) molecules to the appropriate side.
Let's add 6 H2O molecules to the right side:
CrO4^(2-)(aq) + H2O(l) + HSnO2^-(aq) + H+(aq) → CrO2^-(aq) + 6H2O(l) + OH^-(aq) + HSnO3^-(aq)
Now let's count the number of atoms for each element again:
On the left side:
Cr: 1
O: 9
H: 4
Sn: 1
On the right side:
Cr: 1
O: 9
H: 12
Sn: 1
All the atoms are now balanced. Finally, let's write down the coefficients in front of each compound to complete the balanced equation:
1CrO4^(2-)(aq) + 1H2O(l) + 1HSnO2^-(aq) + 1H^+(aq) → 1CrO2^-(aq) + 6H2O(l) + 1OH^-(aq) + 1HSnO3^-(aq)
Now we can obtain the coefficients for the balanced equation:
a = 1, b = 1, c = 1, d = 6
The balanced equation is:
CrO4^(2-)(aq) + H2O(l) + HSnO2^-(aq) + H^+(aq) → CrO2^-(aq) + 6H2O(l) + OH^-(aq) + HSnO3^-(aq)
Thus, the integer abcd representation of the balanced equation is 1116.