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.
Half rxns:
BrO3^- > Br^-
unbalanced oxygens so you would add H2O to balance the oxygen
BrO3^- > Br^- + 3H2O (because you want 3 oxygens on the right as there are on the left)
Then you must balance hydrogens
BrO3^- + 6H^+ > Br^- + 3H2O (you add 6H because there are 6 hydrogens on the right, 3 * 2 = 6)
Now add 6 electrons to the left to cancel out the charge of the 6H^+
BrO3^- + 6H^+ + 6e^- > Br^- + 3H2O
Other half rxn:
Sb^3+ > Sb^5+
You would have to add electrons to the more positive side so the charges on each side are equal
Sb^3+ > Sb^5+ + 2e- (you add 2 electrons because 5-2=3)
Multiply this half reaction by 3 so you can cancel the electrons out from the first half rxn, 2*3=6
Now you have
3Sb^3+ > 3Sb^5+ + 6e^-
Now you can cancel out the electrons and combine each half reaction to get your final answer of:
BrO3^- + 3Sb^3+ + 6H^+ > Br^- + 3Sb^5+ + 3H2O
To complete and balance the equation for the oxidation-reduction titration between BrO3^- and Sb3^+ in acidic solution, follow these steps:
Step 1: Identify the oxidation states of each element in the reaction.
Br: The oxidation state of Br in BrO3^- is +5.
Sb: The oxidation state of Sb in Sb3^+ is +3.
Step 2: Write the unbalanced equation for the reaction:
BrO3^- + Sb3^+ → Br^- + Sb5^+
Step 3: Assign electrons to balance the oxidation and reduction half-reactions.
Oxidation Half-Reaction:
Sb3^+ → Sb5^+
The Sb goes from an oxidation state of +3 to +5, so it loses 2 electrons.
Sb3^+ → Sb5^+ + 2e^-
Reduction Half-Reaction:
BrO3^- → Br^-
The Br goes from an oxidation state of +5 to -1, so it gains 6 electrons.
6BrO3^- + 6e^- → 6Br^-
Step 4: Balance the electrons in both half-reactions.
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to balance the number of electrons transferred.
2(Sb3^+ → Sb5^+ + 2e^-)
3(6BrO3^- + 6e^- → 6Br^-)
Step 5: Combine the balanced half-reactions to form the overall redox equation.
2Sb3^+ + 3BrO3^- + 6H+ → 2Sb5^+ + 6Br^- + 3H2O
Thus, the complete and balanced equation for the reaction in acidic solution is:
2Sb3^+ + 3BrO3^- + 6H+ → 2Sb5^+ + 6Br^- + 3H2O
To balance the equation for the oxidation-reduction reaction in acidic solution, you need to consider the number of atoms of each element on both sides of the equation. Here is the correct balanced equation:
BrO3^- + 3Sb^3+ + 6H+ -> Br^- + 3Sb^5+ + 3H2O
Explanation:
1. Start by balancing the elements other than hydrogen and oxygen. In this case, Sb and Br are the elements to focus on.
2. Balance the element that appears the fewest times on both sides of the equation. In this case, Br appears only once on the reactant side, so we put a coefficient of 1 in front of Br^-.
3. Balance the other element, Sb, by adjusting the coefficients on the other side of the equation. You need three Sb^5+ ions on the product side, so put a coefficient of 3 in front of Sb^3+ on the reactant side.
4. Now, the equation is unbalanced in terms of hydrogen and oxygen atoms. To balance hydrogen, add 6 H+ ions on the reactant side.
5. Finally, balance oxygen atoms by adding 3 H2O molecules on the product side.
6. Double-check that all elements are balanced.
The balanced equation is:
BrO3^- + 3Sb^3+ + 6H+ -> Br^- + 3Sb^5+ + 3H2O