Need help with writing an ionic equation

In summary, when Manganese Dioxide (MnO2) was fused with Sodium Hydroxide (NaOH) and potassium chlorate (KCLO3), a green solid was formed containing K2MnO4 and potassium chloride (KCl). This mixture was then treated with carbon dioxide (CO2) to create an acidic condition and filtered to remove the MnO2, leaving behind a purple filtrate. In standard acidic conditions, the complete ionic equations for the formation of MnO4*2- and MnO4*- are shown. In standard alkaline conditions, the equations are slightly different, with the formation of OH*- instead of H*+. Overall, the addition of CO2 creates an acidic condition,
  • #1
Akeo
1
0
well here is the question:

Manganese Dioxide, MnO2, was fused with Sodium Hydroxide, NaOH, and an oxidising agent potassium chlorate(v), KCLO3. The result was a green solid containing K2MnO4, and potassium chloride, KCL. This mixture was powdered and boiled with water through which carbon dioxide was bubbled for a while. The mixture was then filtered to remove MnO2, and the purple filtrate was carefully evaporated.


In the synthesis above, write the complete ionic equations showing the formation of:
A) MnO4*2-
B)MnO4*-

I put an astrate before the charge so whatever comes after the astrate is the charge on the molecule. And, one more thing, Is the CO2 there just to create an acidic condition.

Here's a little added info.

In standard acid conditions

2MnO4*-(aq) + 2e*- <--> 2MnO4*2-
MnO4*2- (aq)+ 4H*+ + 2e-<--> MnO2(S) + 2H2O(l)

In standard alkaline conditions

2MnO4*-(aq) + 2e*-<-->2MnO4*2-
MnO4*2- (aq)+ 2H2O(l) + 2e-<--> MnO2(S) + 4OH*-
 
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  • #2
A) 2MnO2(s) + 4KClO3(aq) + 8H2O(l) --> 4K2MnO4(s) + 8KCl(aq) + 8OH*-(aq) + 4H*+(aq) B) 2MnO2(s) + 4KClO3(aq) + 8H2O(l) --> 4K2MnO4(s) + 8KCl(aq) + 4O2(g) + 4OH*-(aq)
 
  • #3


The complete ionic equation for the formation of MnO4*2- is:

MnO2(s) + 2NaOH(aq) + 2KClO3(aq) → 2K2MnO4(aq) + 2KCl(aq) + H2O(l)

In this reaction, the MnO2 is oxidized to MnO4*2- by the potassium chlorate and sodium hydroxide. The potassium chloride remains in its ionic form and the water is a spectator ion.

The complete ionic equation for the formation of MnO4*- is:

MnO2(s) + 2NaOH(aq) + 2KClO3(aq) + 2CO2(g) → 2K2MnO4(aq) + 2KCl(aq) + 2H2O(l) + 2CO3*2-(aq)

In this reaction, the carbon dioxide is added to create an acidic condition, which allows for the formation of MnO4*- instead of MnO4*2-. The carbonate ions (CO3*2-) are formed as a byproduct. Again, the potassium chloride remains in its ionic form and the water is a spectator ion.

I hope this helps with your ionic equations. Remember to balance them and include the states of matter for each compound. Good luck!
 

1. What is an ionic equation?

An ionic equation is a chemical equation that only includes the ionic compounds and ions that are involved in a chemical reaction. It shows the transfer of electrons between the reactants and products, rather than the complete molecular structure.

2. How do I write an ionic equation?

To write an ionic equation, you first need to identify the ionic compounds and ions in the reaction. Then, write the balanced chemical equation and separate the ions on either side of the arrow. Finally, cancel out any spectator ions that appear on both sides of the equation to get the final ionic equation.

3. What are spectator ions?

Spectator ions are ions that appear on both sides of a chemical equation and do not participate in the reaction. They can be cancelled out when writing an ionic equation because they do not change in the reaction.

4. Can you give an example of an ionic equation?

Sure, an example of an ionic equation would be the reaction between sodium chloride (NaCl) and silver nitrate (AgNO3) to form silver chloride (AgCl) and sodium nitrate (NaNO3):
NaCl + AgNO3 → AgCl + NaNO3

5. Why is it important to write an ionic equation?

An ionic equation is important because it helps to clearly show the species that are involved in a chemical reaction and the transfer of electrons that occurs. It also allows for the cancellation of spectator ions, making the equation simpler and easier to understand.

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