The quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dissolved in hot, concentrated acid and passed over a reducing agent so that all the antimony is in the form Sb3 (aq). The Sb3 (aq) is completely oxidized by an aqueous solution of BrO3–(aq). Complete and balance the equation for this reaction in acidic solution.
BrO3^- + 3Sb^3-> Br^- + 3Sb^5+
I tried BrO3^- + 3Sb^3+ +6H -> Br^- + 3Sb^5+ +3H2O but it is wrong.
To balance the equation for the oxidation-reduction titration between BrO3–(aq) and Sb3(aq) in acidic solution, you can follow these steps:
1. Write the unbalanced equation:
BrO3–(aq) + Sb3(aq) → Br–(aq) + Sb5+(aq)
2. Balance the atoms involved in the redox reaction, excluding oxygen and hydrogen:
BrO3–(aq) + Sb3(aq) → Br–(aq) + Sb5+(aq)
3. Balance the oxygen atoms by adding water (H2O) molecules:
BrO3–(aq) + Sb3(aq) → Br–(aq) + Sb5+(aq) + H2O(l)
4. Balance the hydrogen atoms by adding H+ ions:
BrO3–(aq) + Sb3(aq) + H+(aq) → Br–(aq) + Sb5+(aq) + H2O(l)
5. Balance the charges by adding electrons (e–):
BrO3–(aq) + Sb3(aq) + 6H+(aq) → Br–(aq) + Sb5+(aq) + 3H2O(l) + 6e–
The balanced equation for the oxidation-reduction titration in acidic solution is:
2BrO3–(aq) + 3Sb3(aq) + 6H+(aq) → 2Br–(aq) + 3Sb5+(aq) + 3H2O(l)
To correctly balance the equation for the oxidation-reduction titration between BrO3–(aq) and Sb3+(aq), you need to take into account that the reaction takes place in an acidic solution. Here's the correctly balanced equation:
3BrO3^-(aq) + 9Sb^3+(aq) + 18H+(aq) → 3Br^-(aq) + 9Sb^5+(aq) + 9H2O(l)
Let's go through the process of balancing the equation step by step.
Step 1: Assign oxidation numbers to each element in the equation:
In this reaction, bromine has an oxidation number of +5 in BrO3^-, and the antimony in Sb3+ has an oxidation number of +3, which increases to +5 in Sb5+.
Step 2: Separate the reaction into two half-reactions:
a) Oxidation half-reaction: Sb3+ → Sb5+
b) Reduction half-reaction: BrO3^- → Br^-
Step 3: Balance the atoms other than hydrogen and oxygen:
a) For the oxidation half-reaction, to balance the antimony atoms, we need 3 Sb3+ on the left side.
b) For the reduction half-reaction, to balance the bromine atoms, we need 3 BrO3^- on the left side.
The equation so far is: 3 BrO3^- + 3 Sb3+ → ? + 3 Sb5+
Step 4: Balance the oxygen atoms by adding water molecules (H2O):
a) There are 9 oxygen atoms in 3 BrO3^- on the left, so we need 9 H2O on the right.
b) No oxygen atoms need to be balanced in the oxidation half-reaction.
The equation becomes: 3 BrO3^- + 3 Sb3+ → ? + 3 Sb5+ + 9 H2O
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+):
a) There are 18 hydrogen atoms in 9 H2O on the right, so we need 18 H+ on the left.
b) To balance the charge in the oxidation half-reaction, we need 18 H+ on the left as well.
The equation becomes: 3 BrO3^- + 3 Sb3+ + 18 H+ → ? + 3 Sb5+ + 9 H2O
Step 6: Balance the charges by adding electrons (e^-):
a) In the oxidation half-reaction, the Sb3+ is oxidized to Sb5+, so it loses 2 electrons. Therefore, we put 2e^- on the right side.
b) There is no charge to be balanced in the reduction half-reaction (BrO3^- to Br^-).
The equation is now: 3 BrO3^- + 3 Sb3+ + 18 H+ → ? + 3 Sb5+ + 9 H2O + 2e^-
Step 7: Ensure the number of electrons in the oxidation and reduction half-reactions is equal:
To equalize the number of electrons, we need to multiply the reduction half-reaction by 2. This way, we will have 4e^- on the right side of the equation.
The equation becomes: 3 BrO3^- + 3 Sb3+ + 18 H+ → ? + 3 Sb5+ + 9 H2O + 4e^-
Step 8: Combine the two half-reactions into a balanced equation:
By combining the half-reactions, we get the final balanced equation:
3 BrO3^- + 9 Sb3+ + 18 H+ → 3 Br^- + 3 Sb5+ + 9 H2O