Use the oxidation number method to balance the following equations by placing coefficients in the blanks. Identify the reducing and oxidizing agents.

_I2+_Na2S2O3-->_Na2S4O6+_NaI

You need to learn how to do these without my help. Here is a site that very good with redox methodology. To give you a hint and help you through the laborious work, I can tell you that BOTH I atoms change from zero on the left to -2 total on the right. S changes from +8 for both S atoms on the left to +10 on the right for all four.

http://www.chemteam.info/Redox/Redox.html

To balance the equation using the oxidation number method, you need to determine the oxidation states for each element in the equation and then adjust the coefficients accordingly. Let's break down the equation and determine the oxidation states for each element:

1. I2 + Na2S2O3 --> Na2S4O6 + NaI

First, let's start with iodine (I). Since it exists as a diatomic molecule (I2), the oxidation state of each iodine atom is 0.

Next, let's consider sodium (Na) in Na2S2O3 and NaI. The oxidation state of sodium is always +1 in compounds.

Moving on to sulfur (S), in Na2S2O3, the total oxidation state of S is -2 since two Na atoms will contribute +2 in total and the overall compound is neutral.

Now, let's consider oxygen (O) in Na2S2O3 and Na2S4O6. In Na2S2O3, each oxygen atom carries an oxidation state of -2. In Na2S4O6, each sulfur atom is assigned an oxidation state of +6, and each oxygen atom is assigned an oxidation state of -2.

Based on the changes in oxidation states, we can propose the following balanced equation:

2I2 + Na2S2O3 --> Na2S4O6 + 4NaI

In the balanced equation, we have:

- 2 iodine atoms on both sides.
- 2 sodium atoms on both sides.
- 1 sulfur atom on both sides.
- 3 oxygen atoms on both sides.

Now, let's identify the reducing and oxidizing agents:

In this reaction, iodine (I2) is reduced from an oxidation state of 0 to -1 in NaI. Therefore, iodine is the oxidizing agent since it causes the reduction of another species.

Sodium thiosulfate (Na2S2O3) is oxidized from an oxidation state of +2 to +6 in Na2S4O6. Therefore, sodium thiosulfate is the reducing agent as it undergoes oxidation and causes the reduction of another species.

To summarize, the balanced equation is:
2I2 + Na2S2O3 --> Na2S4O6 + 4NaI
The reducing agent is Na2S2O3 (sodium thiosulfate), and the oxidizing agent is I2 (iodine).