6K.3D balancing problem?

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Jocelyn Thorp 1A
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Joined: Wed Sep 18, 2019 12:20 am

6K.3D balancing problem?

Postby Jocelyn Thorp 1A » Sun Feb 23, 2020 10:19 pm

Ok so I know that there's a typo for 6K3.D where the one of the products is meant to be 2Cl- rather than Cl2.

For the oxidation reaction I have:
Cl2 + H2O -> HclO + H+ + e-

For the reduction reaction I have:
Cl2 + 2e- -> 2Cl-

so when balancing the electrons you multiply the oxidation reaction by two (to cancel the electrons), and then when you add them together I get:
3Cl2 + 2H2O -> 2HclO + 2Cl- + 2H+

This is definitely not the right answer but I'm not sure where I went wrong. Is anyone able to help me out here?

Chem_Mod
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Re: 6K.3D balancing problem?

Postby Chem_Mod » Sun Feb 23, 2020 10:31 pm

Check your oxidation half-reaction; it's not balanced. You have a 2 chlorines on the reactant side and 1 chlorine on the product side.

805329408
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Joined: Wed Nov 20, 2019 12:21 am

Re: 6K.3D balancing problem?

Postby 805329408 » Sun Feb 23, 2020 11:02 pm

Can someone explain why we can write Cl2 -> 2 Cl- please?

505316964
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Joined: Thu Jul 11, 2019 12:17 am

Re: 6K.3D balancing problem?

Postby 505316964 » Sun Feb 23, 2020 11:07 pm

I believe you write Cl2 --> 2Cl- in the products because this is the only way to go about doing the reduction. If you leave it as Cl2 --> Cl2 nothing is really happening in the reaction and rewriting it is equivalent but now you have a charge to work with.

Jocelyn Thorp 1A
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am

Re: 6K.3D balancing problem?

Postby Jocelyn Thorp 1A » Sun Feb 23, 2020 11:20 pm

so fixing the balancing error I did get the right answer.

also, we can do Cl2 -> 2 Cl- because there was a typo in the book and the equation was supposed to be
Cl2 -> HClO + Cl- where we had to balance the extra H+ and waters.


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