STEP 1 Write a skeleton equation for the reactio.pdf

STEP 1: Write a skeleton equation for the reaction. The skeleton equation for the
reaction on which this titration is based can be written as follows. I3- + S2O32- -----> I- +
S4O62- STEP 2: Assign oxidation numbers to atoms on both sides of the equation. The negative
charge in the I3- ion is formally distributed over the three iodine atoms, which means that the
average oxidation state of the iodine atoms in this ion is -1/3. In the S4O62- ion, the total
oxidation state of the sulfur atoms is +10. The average oxidation state of the sulfur atoms is
therefore +21/2. I3- + S2O32- -----> I- + S4O62- -1/3 +2 -2 -1 +21/2 -2 STEP 3: Determine
which atoms are oxidized and which are reduced. oxired.gif (1729 bytes) STEP 4: Divide the
reaction into oxidation and reduction half-reactions and balance these half-reactions one at a
time. This reaction can be arbitrarily divided into two half-reactions. One half-reaction describes
what happens during oxidation. Oxidation: S2O32- -----> S4O62- +2 +21/2 The other describes
the reduction half of the reaction. Reduction: I3- -----> I- -1/3 -1 It doesn\'t matter which half-
reaction we balance first, so let\'s start with the reduction half-reaction. Our goal is to balance
this half-reaction in terms of both charge and mass. It seems reasonable to start by balancing the
number of iodine atoms on both sides of the equation. Reduction: I3- -----> 3 I- We then balance
the charge by noting that two electrons must be added to an I3- ion to produce 3 I- ions,
Reduction: I3- + 2 e- -----> 3 I- as can be seen from the Lewis structures of these ions shown in
the figure below. We now turn to the oxidation half-reaction. The Lewis structures of the
starting material and the product of this half-reaction suggest that we can get an S4O62- ion by
removing two electrons from a pair of S2O32- ions, as shown in the figure below. Oxidation: 2
S2O32- -----> S4O62- + 2 e- STEP 5: Combine these half-reactions so that electrons are neither
created nor destroyed. Two electrons are given off in the oxidation half-reaction and two
electrons are picked up in the reduction half-reaction. We can therefore obtain a balanced
chemical equation by simply combining these half-reactions. (2 S2O32------>S4O62- + 2 e-) +
(I3- + 2 e- ----->3 I-) I3- + 2 S2O32- ----->3 I- + S4O62- STEP 6: Balance the remainder of the
equation by inspection, if necessary. Since the overall equation is already balanced in terms of
both charge and mass, we simply introduce the symbols describing the states of the reactants and
products. I3-(aq) + 2 S2O32-(aq) ----->3 I-(aq) + S4O62-(aq)
Solution
STEP 1: Write a skeleton equation for the reaction. The skeleton equation for the
reaction on which this titration is based can be written as follows. I3- + S2O32- -----> I- +
S4O62- STEP 2: Assign oxidation numbers to atoms on both sides of the equation. The negative
charge in the I3- ion is formally distributed over the three iodine atoms, which means that the
average oxidatio.

1. STEP 1: Write a skeleton equation for the reaction. The skeleton equation for the
reaction on which this titration is based can be written as follows. I3- + S2O32- -----> I- +
S4O62- STEP 2: Assign oxidation numbers to atoms on both sides of the equation. The negative
charge in the I3- ion is formally distributed over the three iodine atoms, which means that the
average oxidation state of the iodine atoms in this ion is -1/3. In the S4O62- ion, the total
oxidation state of the sulfur atoms is +10. The average oxidation state of the sulfur atoms is
therefore +21/2. I3- + S2O32- -----> I- + S4O62- -1/3 +2 -2 -1 +21/2 -2 STEP 3: Determine
which atoms are oxidized and which are reduced. oxired.gif (1729 bytes) STEP 4: Divide the
reaction into oxidation and reduction half-reactions and balance these half-reactions one at a
time. This reaction can be arbitrarily divided into two half-reactions. One half-reaction describes
what happens during oxidation. Oxidation: S2O32- -----> S4O62- +2 +21/2 The other describes
the reduction half of the reaction. Reduction: I3- -----> I- -1/3 -1 It doesn\'t matter which half-
reaction we balance first, so let\'s start with the reduction half-reaction. Our goal is to balance
this half-reaction in terms of both charge and mass. It seems reasonable to start by balancing the
number of iodine atoms on both sides of the equation. Reduction: I3- -----> 3 I- We then balance
the charge by noting that two electrons must be added to an I3- ion to produce 3 I- ions,
Reduction: I3- + 2 e- -----> 3 I- as can be seen from the Lewis structures of these ions shown in
the figure below. We now turn to the oxidation half-reaction. The Lewis structures of the
starting material and the product of this half-reaction suggest that we can get an S4O62- ion by
removing two electrons from a pair of S2O32- ions, as shown in the figure below. Oxidation: 2
S2O32- -----> S4O62- + 2 e- STEP 5: Combine these half-reactions so that electrons are neither
created nor destroyed. Two electrons are given off in the oxidation half-reaction and two
electrons are picked up in the reduction half-reaction. We can therefore obtain a balanced
chemical equation by simply combining these half-reactions. (2 S2O32------>S4O62- + 2 e-) +
(I3- + 2 e- ----->3 I-) I3- + 2 S2O32- ----->3 I- + S4O62- STEP 6: Balance the remainder of the
equation by inspection, if necessary. Since the overall equation is already balanced in terms of
both charge and mass, we simply introduce the symbols describing the states of the reactants and
products. I3-(aq) + 2 S2O32-(aq) ----->3 I-(aq) + S4O62-(aq)

Solution
STEP 1: Write a skeleton equation for the reaction. The skeleton equation for the
reaction on which this titration is based can be written as follows. I3- + S2O32- -----> I- +
S4O62- STEP 2: Assign oxidation numbers to atoms on both sides of the equation. The negative
charge in the I3- ion is formally distributed over the three iodine atoms, which means that the
average oxidation state of the iodine atoms in this ion is -1/3. In the S4O62- ion, the total
oxidation state of the sulfur atoms is +10. The average oxidation state of the sulfur atoms is

2. therefore +21/2. I3- + S2O32- -----> I- + S4O62- -1/3 +2 -2 -1 +21/2 -2 STEP 3: Determine
which atoms are oxidized and which are reduced. oxired.gif (1729 bytes) STEP 4: Divide the
reaction into oxidation and reduction half-reactions and balance these half-reactions one at a
time. This reaction can be arbitrarily divided into two half-reactions. One half-reaction describes
what happens during oxidation. Oxidation: S2O32- -----> S4O62- +2 +21/2 The other describes
the reduction half of the reaction. Reduction: I3- -----> I- -1/3 -1 It doesn\'t matter which half-
reaction we balance first, so let\'s start with the reduction half-reaction. Our goal is to balance
this half-reaction in terms of both charge and mass. It seems reasonable to start by balancing the
number of iodine atoms on both sides of the equation. Reduction: I3- -----> 3 I- We then balance
the charge by noting that two electrons must be added to an I3- ion to produce 3 I- ions,
Reduction: I3- + 2 e- -----> 3 I- as can be seen from the Lewis structures of these ions shown in
the figure below. We now turn to the oxidation half-reaction. The Lewis structures of the
starting material and the product of this half-reaction suggest that we can get an S4O62- ion by
removing two electrons from a pair of S2O32- ions, as shown in the figure below. Oxidation: 2
S2O32- -----> S4O62- + 2 e- STEP 5: Combine these half-reactions so that electrons are neither
created nor destroyed. Two electrons are given off in the oxidation half-reaction and two
electrons are picked up in the reduction half-reaction. We can therefore obtain a balanced
chemical equation by simply combining these half-reactions. (2 S2O32------>S4O62- + 2 e-) +
(I3- + 2 e- ----->3 I-) I3- + 2 S2O32- ----->3 I- + S4O62- STEP 6: Balance the remainder of the
equation by inspection, if necessary. Since the overall equation is already balanced in terms of
both charge and mass, we simply introduce the symbols describing the states of the reactants and
products. I3-(aq) + 2 S2O32-(aq) ----->3 I-(aq) + S4O62-(aq)