Balance Redox Rxn: KBrO3 + NH3OHCL

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In summary, the most likely product of the given reaction is NO3-, and the balanced equation using the half-reaction method is KBrO3 + NH3OHCl + H2O --> KBr + NO3- + 2H+ + Cl-.
  • #1
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Homework Statement



“From knowledge of oxidation states of nitrogen in the product as determined experimentally, and the oxidation state of nitrogen in each species in the list of possible products, select the most likely product of the following:
….NO3- (I found this to be the product)
Write the balanced REDOX equation for the reaction between bromate and hydroxylammonium using the ½ reaction method.”

So basically I have to balance the chemical equation using the ‘half rxn method’ in order to get NO3- but I’m not sure if my final balanced equation is right and if I’ve done the oxidation half and reduction half reactions correctly.


Homework Equations



KBrO3 + NH3OHCL ------> Br- + Some nitrogen containing product (I found the product to be NO3-)

In aqueous solution


The Attempt at a Solution



I made K and Cl spectator ions

Not writing the ½ reactions, I got this for the final equation:

BrO3- + H2O + NH3OH+ --> Br- + NO3- + 2H3O+

Problem is that I’m having trouble putting the spectator ions (Cl and K) back into the equation. Could someone show me how to add Cl back in and how to properly do the oxidation/reduction halves?
 
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  • #2



Hello, based on the information given, it seems that the most likely product is NO3-. To balance the equation using the half-reaction method, you first need to identify the oxidation state of each element in the reactants and products. In this case, nitrogen has an oxidation state of +5 in NO3- and +3 in NH3OH+. This means that it is being reduced from +5 to +3 in the reaction.

Next, you need to balance the number of atoms on each side of the equation. In the oxidation half-reaction, you have 1 N atom on the reactant side and 3 N atoms on the product side. To balance this, you can add 2H2O on the reactant side, which will give you 2 N atoms on both sides.

In the reduction half-reaction, you have 1 N atom on the reactant side and 1 N atom on the product side. To balance this, you can add 2H+ on the product side, which will give you 2 N atoms on both sides.

Now, you can combine the half-reactions and cancel out any common terms. This will give you the balanced equation:

BrO3- + NH3OH+ + H2O --> Br- + NO3- + 2H+

To add the spectator ions back in, you can simply write them on both sides of the equation:

KBrO3 + NH3OHCl + H2O --> KBr + NO3- + 2H+ + Cl-

I hope this helps! Let me know if you have any further questions.
 
  • #3




First, let's determine the oxidation states of the elements in the given reactants and products:

- KBrO3: K has an oxidation state of +1, Br has an oxidation state of +5, and O has an oxidation state of -2
- NH3OHCL: N has an oxidation state of -3, H has an oxidation state of +1, Cl has an oxidation state of -1
- Br-: Br has an oxidation state of -1
- NO3-: N has an oxidation state of +5, O has an oxidation state of -2

Based on the given information, it is likely that the product of this redox reaction is NO3-. To balance the equation using the half reaction method, we need to first write out the oxidation and reduction half reactions separately.

Oxidation half reaction: NH3OHCL --> NO3- + 2H+ + 3e-
Reduction half reaction: BrO3- + 6H+ + 6e- --> Br- + 3H2O

Next, we need to balance the number of atoms and charge on each side of the half reactions. In the oxidation half reaction, there are 3 H atoms on the right side, so we need to add 3 H2O molecules to the left side to balance it. We also need to add 3 electrons to the right side to balance the charge. This gives us the balanced oxidation half reaction:

NH3OHCL + 3H2O --> NO3- + 2H+ + 3e-

Similarly, in the reduction half reaction, there are 6 H atoms and 6 H+ ions on the left side, so we need to add 6 H2O molecules to the right side to balance it. We also need to add 6 electrons to the left side to balance the charge. This gives us the balanced reduction half reaction:

BrO3- + 6H+ + 6e- --> Br- + 3H2O

Now, we can combine the two half reactions to form the overall balanced redox equation:

NH3OHCL + BrO3- + 3H2O --> NO3- + Br- + 3H2O + 2H+

Simplifying, we get the final balanced equation:

NH3OHCL + BrO3- --> NO3- +
 

1. What is the balanced redox reaction for KBrO3 + NH3OHCl?

The balanced redox reaction for KBrO3 + NH3OHCl is 3KBrO3 + 3NH3OHCl → 3KBr + N2 + 6H2O.

2. What is the oxidation state of each element in KBrO3 + NH3OHCl?

The oxidation state of K in KBrO3 is +1, Br in KBrO3 is +5, O in KBrO3 is -2, N in NH3OHCl is -3, H in NH3OHCl is +1, and Cl in NH3OHCl is -1.

3. How do you determine which element is oxidized and which is reduced in KBrO3 + NH3OHCl?

In KBrO3 + NH3OHCl, K is reduced from +1 to 0, while Br is oxidized from +5 to 0. N is reduced from -3 to 0, while H is oxidized from +1 to +2. Therefore, K and N are the reducing agents, and Br and H are the oxidizing agents.

4. How many electrons are transferred in the redox reaction of KBrO3 + NH3OHCl?

In the redox reaction of KBrO3 + NH3OHCl, a total of 6 electrons are transferred. 3 electrons are transferred from K to Br, and 3 electrons are transferred from N to H.

5. What is the overall change in oxidation state for the redox reaction of KBrO3 + NH3OHCl?

The overall change in oxidation state for the redox reaction of KBrO3 + NH3OHCl is 0. K and N both change from +1 and -3 to 0, while Br and H both change from +5 and +1 to 0 and +2, respectively.

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