When the following reaction is balanced in acid solution, the coefficient of H+(aq) will be:
Cr2O72-(aq) + Sn2+(aq) --> Cr3+(aq) + Sn4+(aq)
28H+ 3Sn2+ + 2(Cr2O7)2- ------> 3Sn4+ + 4Cr3+ + 14H2O
To balance the given chemical reaction in an acidic solution, we need to add hydrogen ions (H+) in the form of acidic water (H2O) to the appropriate side of the reaction. Here's how you can balance the reaction:
1. Start by assigning coefficients to the reactants and products that have not been balanced. In this case, Cr2O72-(aq) and Sn2+(aq) are the reactants, and Cr3+(aq) and Sn4+(aq) are the products.
Cr2O72-(aq) + Sn2+(aq) --> Cr3+(aq) + Sn4+(aq)
2. Balance the elements that appear in multiple compounds first. In this reaction, chromium (Cr) and oxygen (O) appear in multiple compounds.
To balance the chromium (Cr) atoms, add coefficients as needed. In this case, since there are two chromium (Cr) atoms on the left side and three on the right side, we can balance it by adding a coefficient of 2 in front of Cr3+(aq):
Cr2O72-(aq) + Sn2+(aq) --> 2 Cr3+(aq) + Sn4+(aq)
Now, let's balance the oxygen atoms (O). Count the number of oxygen atoms on each side of the reaction equation. We have 14 oxygen atoms on the left side and only 6 on the right side. To balance them, we add water (H2O) to the reaction:
Cr2O72-(aq) + Sn2+(aq) --> 2 Cr3+(aq) + Sn4+(aq) + H2O
However, when we add water, we also increase the number of hydrogen (H) atoms. We need to balance this by adding hydrogen ions (H+) to both sides of the reaction equation. The number of H+ ions we add will be equal to the number of oxygen atoms we added with water. In this case, we add 8 H+ ions on both sides:
Cr2O72-(aq) + Sn2+(aq) + 8H+(aq) --> 2 Cr3+(aq) + Sn4+(aq) + H2O + 8H+(aq)
3. Finally, check that all elements and charges are balanced. Count the number of atoms on each side of the equation to ensure that they are equal. Also, check that the charges are balanced.
On the left side, we have 2 Cr, 7 O, 8 H, and 1 Sn. On the right side, we have 2 Cr, 7 O, 8 H, and 1 Sn. All elements are balanced.
The coefficient of H+(aq) is 8.
So, when the given reaction is balanced in acid solution, the coefficient of H+(aq) is 8.
I understand you are supposed to separate into half reactions and balance using H20 H+ and e- however, when i start doing that I get confused halfway through.
this is as far as I have gotten:
4e- + 7H+ + Cr2O72- = 2Cr3+ +7H20
2Sn2+ = 2Sn4+ +4e-
then cancel electrons...is this right?
اسيل
4e- + 7H+ + Cr2O72- = 2Cr3+ +7H20
Since you didn't write the entire equation I can't tell if everything is separated or not; however, the Cr half reaction is separated properly. The trouble you have in balancing is because you have assigned the wrong number of electrons. Cr in Cr2O7^2- is +12 for the two and +6 on the right for the two. That makes all of the other numbers wrong.
6e + Cr2O^2- ==> 2Cr^+3
Now you count charges on the left (-8) and on the right(+6) and add 14 H^+ on the left.
14H^+ + 6e + Cr2O7^2- ==> 2Cr^3+
Now add 7H2O on the right to balancae.
14H^+ + 6e + Cr2O7^2- ==> 2Cr^3+ + 7H2O.
I don't know if you've separated the Sn half cell properly or not.