I2 + 5 Cl2 + 6 H2O 2 HIO3 + 10 HCl
What is the reducing agent? In your answer give the element symbol and oxidation state.
Where's the arrow? How do you know where the reactants stop and products start without an arrow?
I changes from zero to +5. the right. Cl changes from zero to -1.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
The substance oxidized is the reducing agent.
The substance reduced is the oxidizing agent.
To determine the reducing agent in the given equation, we need to find the element that undergoes a decrease in its oxidation state. Here's how you can identify it:
1. Start by assigning oxidation states to all the elements in the equation based on their electronegativities and known oxidation rules. In this case, let's consider chlorine (Cl) and iodine (I).
2. Chlorine typically has an oxidation state of -1 in compounds, and oxygen (O) has an oxidation state of -2. Since there are five Cl2 molecules, the total oxidation number for chlorine would be -10.
3. Oxygen also typically has an oxidation state of -2, and there are six H2O molecules, so the total oxidation number for oxygen is -12.
4. Considering that iodine (I) is present on both sides of the equation, it is necessary to check if there has been any change in its oxidation state.
5. Comparing the oxidation state of iodine on the left side of the equation (I2) to that on the right side (HIO3), we observe that iodine's oxidation state has increased from 0 in I2 to +5 in HIO3.
6. This increase in oxidation state indicates that iodine has undergone oxidation, meaning it has lost electrons. Therefore, the reducing agent would be the element that has caused this oxidation, which is chlorine (Cl).
7. To identify the oxidation state of chlorine, we can compare Cl2 on the left side of the equation to the product on the right side, which is 10 HCl. Since there are 10 HCl molecules, each Cl atom must have an oxidation state of -1.
So, the reducing agent in this equation is chlorine (Cl) with an oxidation state of -1.