Question
Balance the reaction I^- (aq) + S2O8^2-(aq)=I^-3 + SO2(g) in acid solution.
Answers
GPT 3.5
To balance the reaction I^- (aq) + S2O8^2- (aq) = I^-3 + SO2 (g) in acid solution, follow these steps:
1. Determine the oxidation states of each element in the reaction. In this case, Iodine (I) has a -1 oxidation state and Sulfur (S) has a +6 oxidation state.
2. Divide the reaction into two half-reactions, one for the reduction and one for the oxidation.
a. Reduction half-reaction: I^- (aq) = I^-3
b. Oxidation half-reaction: S2O8^2- (aq) = SO2 (g)
3. Balance the atoms other than H and O in each half-reaction.
a. For the reduction half-reaction, there is only one iodine (I) on both sides, so it is already balanced.
b. For the oxidation half-reaction, there are two sulfur (S) atoms on the left side and one sulfur (S) atom on the right side. Place a coefficient of 2 in front of SO2 (g) to balance the number of sulfur (S) atoms.
4. Balance the charge in each half-reaction by adding electrons (e^-) to the more positive side of the reaction.
a. For the reduction half-reaction, the oxidation state of iodine (I) is increasing from -1 to -3. To balance the charge, we need to add 2 electrons (2e^-) to the left side.
b. For the oxidation half-reaction, there is no need to balance the charge as the oxidation state of sulfur (S) remains the same.
5. Balance the number of electrons transferred between the two half-reactions.
To balance the number of electrons, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3. This gives:
a. Reduction half-reaction: 2I^- (aq) = I^-3 + 2e^-
b. Oxidation half-reaction: 3S2O8^2- (aq) = 3SO2 (g)
6. Add the two balanced half-reactions together.
2I^- (aq) + 3S2O8^2- (aq) = I^-3 + 2e^- + 3SO2 (g)
7. Cancel out the electrons (e^-) that appear on both sides of the equation.
2I^- (aq) + 3S2O8^2- (aq) = I^-3 + 3SO2 (g)
Now the reaction is balanced in acid solution.
1. Determine the oxidation states of each element in the reaction. In this case, Iodine (I) has a -1 oxidation state and Sulfur (S) has a +6 oxidation state.
2. Divide the reaction into two half-reactions, one for the reduction and one for the oxidation.
a. Reduction half-reaction: I^- (aq) = I^-3
b. Oxidation half-reaction: S2O8^2- (aq) = SO2 (g)
3. Balance the atoms other than H and O in each half-reaction.
a. For the reduction half-reaction, there is only one iodine (I) on both sides, so it is already balanced.
b. For the oxidation half-reaction, there are two sulfur (S) atoms on the left side and one sulfur (S) atom on the right side. Place a coefficient of 2 in front of SO2 (g) to balance the number of sulfur (S) atoms.
4. Balance the charge in each half-reaction by adding electrons (e^-) to the more positive side of the reaction.
a. For the reduction half-reaction, the oxidation state of iodine (I) is increasing from -1 to -3. To balance the charge, we need to add 2 electrons (2e^-) to the left side.
b. For the oxidation half-reaction, there is no need to balance the charge as the oxidation state of sulfur (S) remains the same.
5. Balance the number of electrons transferred between the two half-reactions.
To balance the number of electrons, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3. This gives:
a. Reduction half-reaction: 2I^- (aq) = I^-3 + 2e^-
b. Oxidation half-reaction: 3S2O8^2- (aq) = 3SO2 (g)
6. Add the two balanced half-reactions together.
2I^- (aq) + 3S2O8^2- (aq) = I^-3 + 2e^- + 3SO2 (g)
7. Cancel out the electrons (e^-) that appear on both sides of the equation.
2I^- (aq) + 3S2O8^2- (aq) = I^-3 + 3SO2 (g)
Now the reaction is balanced in acid solution.