6K.3 Part D
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6K.3 Part D
The answer key converts Cl2 on the products side of the equation to 2Cl-. Why does it do this? How would I know to do this is a similar problem?
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Re: 6K.3 Part D
This one was very confusing for me as well, but I think since Cl2 has to serve as both the oxidizing and reducing agent, if Cl is getting oxidized in Cl2 --> HClO, then the Cl2 also needs to get reduced somehow. The easiest way for this to occur would be to make the other half reaction Cl2 --> 2 Cl-.
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Re: 6K.3 Part D
Since Chlorine is the only non-oxygen/hydrogen element involved in this reaction, it can be assumed that it is both the oxidizing and reducing agent. The oxidation half reaction is relatively simple to determine based upon the rules described in Toolbox 6K.1. Since you know that Cl2 is also the oxidizing agent, you know that you need to come up with a half reaction in which it is reduced. Cl2+2e--->2Cl- gets the job done!
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- Posts: 110
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Re: 6K.3 Part D
Ruby Tang 2J wrote:This one was very confusing for me as well, but I think since Cl2 has to serve as both the oxidizing and reducing agent, if Cl is getting oxidized in Cl2 --> HClO, then the Cl2 also needs to get reduced somehow. The easiest way for this to occur would be to make the other half reaction Cl2 --> 2 Cl-.
Okay thank you. That makes more sense.
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