The redox reaction:
O3SOOSO3^2− + Cr^3+ -> HSO4^− + Cr2O7^2−
is carried out in aqueous acidic solution. Which elements are undergoing changes in oxidation state?
I thought it would be S and Cr. For the balanced equation, I got 3O3SOOSO3+Cr^3+ + 13H^+ -> 3HSO4^- + Cr2O7^2- + 5H2O , but it's wrong and I don't know what I did. Thanks in advance.
S and Cr changing sounds logical although I didn't go through it. One thing I spotted right off the bat. You have 1 Cr^+3 on the left and Cr2O7^-2 on the right. The Cr atoms don't balance. Theck the others for good measure. Also check to see that the charges balance; don't think so. You have 6 S on the left and only 3 on the right.
To determine which elements are undergoing changes in oxidation state in the given redox reaction, we need to assign oxidation numbers to the elements in each compound before and after the reaction.
Let's analyze the given reaction step by step:
Given redox reaction: O3SOOSO3^2− + Cr^3+ -> HSO4^− + Cr2O7^2−
1. Assign oxidation numbers:
Start by determining the oxidation states of each element in the compounds before and after the reaction.
O3SOOSO3^2−:
Since oxygen (O) is usually assigned an oxidation state of -2 in most compounds, and there are four oxygen atoms, the total oxidation state contributed by oxygen is -8. The sulfur (S) in the compound has an oxidation state of +6 to balance the negative oxidation state of oxygen. Therefore, the overall oxidation state of S in O3SOOSO3^2− is +6.
Cr^3+:
The oxidation state of chromium (Cr) in its elemental form is 0. In this case, it is given that Cr is in the +3 oxidation state (Cr^3+).
HSO4^−:
As sulfur (S) is more electronegative than hydrogen (H), it is assigned an oxidation state of +6, which balances the -2 oxidation state of oxygen. Therefore, the oxidation state of S in HSO4^− is +6.
Cr2O7^2−:
Again, assuming oxygen has an oxidation state of -2, and there are seven oxygen atoms, the total oxidation state contributed by oxygen is -14. Chromium (Cr) is assigned a variable oxidation state of x in this compound. The overall oxidation state of this compound is -2. By equating the sum of oxidation states to the overall charge, we can solve for x to get the oxidation state of Cr in Cr2O7^2−.
2. Determine the changes in oxidation state:
Now that we have assigned oxidation states to each element in both the reactants and products, we can determine the changes in oxidation state.
For sulfur (S):
The oxidation state of S changes from +6 in O3SOOSO3^2− to +6 in HSO4^−. Therefore, there is no change in oxidation state for S in this reaction.
For chromium (Cr):
The oxidation state of Cr changes from +3 in Cr^3+ to a different oxidation state in Cr2O7^2−. By using the equation:
2x + 7(-2) = -2 (equating the sum of oxidation states to the overall charge),
Solving this equation gives x = +6. Therefore, the oxidation state of Cr changes from +3 to +6 in this reaction.
In summary, the elements undergoing changes in oxidation state in the given redox reaction are chromium (Cr) and sulfur (S).
Regarding your balanced equation attempt, let's balance it step by step:
Given reaction: O3SOOSO3^2− + Cr^3+ + H+ -> HSO4^− + Cr2O7^2− + H2O
Balancing the number of atoms of each element:
Cr: The Cr atoms are already balanced on both sides.
O: There are 3 O atoms on the reactant side and 14 O atoms on the product side. To balance O, add 11 H2O on the reactant side.
H: There are 1 H atom on the reactant side and 17 H atoms on the product side. To balance H, add 16 H+ on the reactant side.
S: The S atoms are already balanced on both sides.
The balanced equation is:
3O3SOOSO3^2− + Cr^3+ + 11H+ -> 3HSO4^− + Cr2O7^2− + 11H2O
Remember to always check if the charges are balanced as well.
I hope this helps! Let me know if you have any further questions.