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Can someone please explain how this is a redox reaction in terms of oxidation states ?

CuO + 2HCl —> CuCl2 + H2O
Original post by 513N4_:D
Can someone please explain how this is a redox reaction in terms of oxidation states ?

CuO + 2HCl —> CuCl2 + H2O

The Oxygen in the CuO has oxidation state of -2. In H2O, oxygen was oxidised and has an oxidation no. of 0.
The Hydrogen in HCl has oxidation state of +1. In H2O, hydrogen was reduced and has oxidation no. of 0.
If you're wondering why 0, water is a liquid
Original post by yesusername
The Oxygen in the CuO has oxidation state of -2. In H2O, oxygen was oxidised and has an oxidation no. of 0.
The Hydrogen in HCl has oxidation state of +1. In H2O, hydrogen was reduced and has oxidation no. of 0.
If you're wondering why 0, water is a liquid


Thank you !
Original post by 513N4_:D
Can someone please explain how this is a redox reaction in terms of oxidation states ?

CuO + 2HCl —> CuCl2 + H2O

It is not redox. There are no changes to any of the oxidation numbers of any of the elements.
Original post by yesusername
The Oxygen in the CuO has oxidation state of -2. In H2O, oxygen was oxidised and has an oxidation no. of 0.
The Hydrogen in HCl has oxidation state of +1. In H2O, hydrogen was reduced and has oxidation no. of 0.
If you're wondering why 0, water is a liquid

Where are you getting these oxidation numbers from? In H2O, the oxidation number of oxygen is still -2. The hydrogen's oxidation number is also +1. There is no redox reaction, as Pigster has rightly said.
Original post by yesusername
The Oxygen in the CuO has oxidation state of -2.


You should have stopped there.
Original post by Pigster
It is not redox. There are no changes to any of the oxidation numbers of any of the elements.

Oh okay.
That is what I originally thought but it came up in a multiple choice question: Which reaction does NOT represent a redox reaction?

I selected that equation but it turned out to be wrong and I was confused about how it would be a redox reaction. Guess it was just a mistake in the question then. Thanks!

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