Consider the reaction in aqueous solution:
6 HCl + 2 KMnO4 + 5 H2O2 ---> 5 O2 + 2 MnCl2 + 2 KCl + 8 H2O
1.) Write the total ionic equation.
2.) Write the net ionic equation.
3.) For each species (molecule or ion) in the net ionic equation, assign oxidation numbers to H, O, and Mn.
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To write the total ionic equation, we need to break down all the reactants and products into their respective ions. In this reaction, we have ionic compounds and some molecular compounds that will dissociate into ions in aqueous solution.
1.) Total ionic equation:
First, let's write out the balanced equation:
6 HCl + 2 KMnO4 + 5 H2O2 ---> 5 O2 + 2 MnCl2 + 2 KCl + 8 H2O
Now, let's break down the reactants that dissociate into ions:
6 H+ + 6 Cl- + 2 K+ + 2 MnO4- + 10 H2O2 ---> 5 O2 + 2 Mn2+ + 2 Cl- + 2 K+ + 8 H2O
Since the Cl- and K+ ions appear unchanged on both sides of the equation, we can cancel them out:
6 H+ + 2 MnO4- + 10 H2O2 ---> 5 O2 + 2 Mn2+ + 8 H2O
This is the total ionic equation.
2.) Net ionic equation:
To obtain the net ionic equation, we need to exclude the spectator ions, which are the ones that appear on both sides of the equation and do not participate in the reaction.
The spectator ions in this reaction are H+ and Mn2+. They remain unaltered throughout the reaction. So, we can remove them from the equation:
10 H2O2 ---> 5 O2 + 8 H2O
This is the net ionic equation.
3.) Assigning oxidation numbers:
The oxidation numbers help us determine the gain or loss of electrons by each element in the reaction. In this case, we need to assign oxidation numbers to H, O, and Mn in the net ionic equation.
In H2O2, each H atom has an oxidation number of +1 since it's bonded to a more electronegative oxygen atom, which has an oxidation number of -1.
In O2, the oxidation number of each O atom is 0 since it is a diatomic molecule.
In Mn2+, since it is listed as an ion, its oxidation number is +2.
Overall, we have:
H2O2: H = +1, O = -1
O2: O = 0
Mn2+: Mn = +2