Net Ionic Equation Lab Report
Writing Half Reaction and Net Ionic Equation
Going back to our first concept or topic, we’ve learn how to identify the Oxidizing and Reducing agent and the Oxidation as well as the Reduction. In this part of the Redox reaction or in this step, it is very important that you already know the Oxidizing and Reducing agent for us to find the half reaction.
In this step or part, we will also learn how to balance using the addition of electrons in both sides if necessary. After this step we can now write the net ionic equation.
There are guidelines that will help you to write the step by step in getting the net ionic equation.
+ Spectator Ion- You cannot change the charges of the elements but happen to be present in the chemical equation. So …show more content…
4th Step- Balance charges
Ox Hrxn: Mg¬0 Mg+2 + 2e-
Red Hrxn: O20 + 2e- 2O-2
In balancing this equation, you will add electron to reactant or product side that will make it equal.
5th Step- Cancel equal number of electrons
Ox Hrxn: Mg¬0 Mg+2 + 2e- ~ Mg¬0 Mg+2
Red Hrxn: O20 + 2e- 2O-2 ~ O20 …show more content…
Examples
1. Fe0 + O20 Fe23 O3-2
As what we’ve discuss, we will just follow the guidelines for us to be able to answer the chemical equation.
Ox Hrxn: Fe0 Fe23 (Reducing Agent)
Red Hrxn: O20 O3-2 (Oxidizing Agent)
Of course you will right for the half reaction first and determine the reducing and oxidizing agent before balancing the number of atoms and charges.
Ox Hrxn: 2 Fe0 Fe23 +6e-
Red Hrxn: 3 O20 +12e- 2O3-2
We balance the number of atoms in Iron (Fe) because in the product side there are 2 Fe that’s why I add up the coefficient of 2 in the reactant side. Where did we get the 6 electrons that are being added to the Iron in product side? By simply multiplying the number of Iron to its oxidation number. It equals to 6 that’s why we will add 6 so that it will be equal. Same to do with the reduction reaction, balance number of atoms and charges by adding electrons.
Ox Hrxn: (2 Fe0 Fe23 +6e- ) 2 ~ 4 Fe02 Fe23 + 12e-
Red Hrxn: 3 O20 +12e- 2O3-2
So in this part, the oxidation reaction we multiply it by 2 so we can cancel the 12e- in reduction
Going back to our first concept or topic, we’ve learn how to identify the Oxidizing and Reducing agent and the Oxidation as well as the Reduction. In this part of the Redox reaction or in this step, it is very important that you already know the Oxidizing and Reducing agent for us to find the half reaction.
In this step or part, we will also learn how to balance using the addition of electrons in both sides if necessary. After this step we can now write the net ionic equation.
There are guidelines that will help you to write the step by step in getting the net ionic equation.
+ Spectator Ion- You cannot change the charges of the elements but happen to be present in the chemical equation. So …show more content…
4th Step- Balance charges
Ox Hrxn: Mg¬0 Mg+2 + 2e-
Red Hrxn: O20 + 2e- 2O-2
In balancing this equation, you will add electron to reactant or product side that will make it equal.
5th Step- Cancel equal number of electrons
Ox Hrxn: Mg¬0 Mg+2 + 2e- ~ Mg¬0 Mg+2
Red Hrxn: O20 + 2e- 2O-2 ~ O20 …show more content…
Examples
1. Fe0 + O20 Fe23 O3-2
As what we’ve discuss, we will just follow the guidelines for us to be able to answer the chemical equation.
Ox Hrxn: Fe0 Fe23 (Reducing Agent)
Red Hrxn: O20 O3-2 (Oxidizing Agent)
Of course you will right for the half reaction first and determine the reducing and oxidizing agent before balancing the number of atoms and charges.
Ox Hrxn: 2 Fe0 Fe23 +6e-
Red Hrxn: 3 O20 +12e- 2O3-2
We balance the number of atoms in Iron (Fe) because in the product side there are 2 Fe that’s why I add up the coefficient of 2 in the reactant side. Where did we get the 6 electrons that are being added to the Iron in product side? By simply multiplying the number of Iron to its oxidation number. It equals to 6 that’s why we will add 6 so that it will be equal. Same to do with the reduction reaction, balance number of atoms and charges by adding electrons.
Ox Hrxn: (2 Fe0 Fe23 +6e- ) 2 ~ 4 Fe02 Fe23 + 12e-
Red Hrxn: 3 O20 +12e- 2O3-2
So in this part, the oxidation reaction we multiply it by 2 so we can cancel the 12e- in reduction