Hi,
For this problem:
Part b- Where does the 3H+ come from? I thought that since there is only 1 H2O, it would be "2 H+" instead?
Part c- When you are balancing the 1/2 reaction, you don't need to add H2O? You can just add H+ if it balances out (see H2S--> S + 2H+)?
Part d- I was just confused in general how the 1/2 reactions were formed.
Exercise 6K Question 3
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Re: Exercise 6K Question 3
For part B, you need to balance the H that is also on H2So3, so you would need 3H+
For part C, I believe you only really add water when balancing oxygens. Its an acidic solution so there is protons available to balance the reaction.
For part D, I am not entirely sure but I guess that because Cl can also be reduced and oxidized with two electrons one of the half reactions is
and the other is . I am not particularly sure why there is a Cl on both sides but I am guessing that is to indicate that Cl2 is being reduced and oxidized.
For part C, I believe you only really add water when balancing oxygens. Its an acidic solution so there is protons available to balance the reaction.
For part D, I am not entirely sure but I guess that because Cl can also be reduced and oxidized with two electrons one of the half reactions is
and the other is . I am not particularly sure why there is a Cl on both sides but I am guessing that is to indicate that Cl2 is being reduced and oxidized.
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Re: Exercise 6K Question 3
For part C, you're right in saying that you don't need to add h2o for the half reaction. Unless there is already an oxygen in the equation that you need to balance on the other side, you can't add additional elements (the oxygen in this case) that aren't already present in the equation because that would change the chemical makeup of the reaction itself. This is how I interpreted it, but feel free to correct or add anything.
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Re: Exercise 6K Question 3
Thanks for the explanations! I was also a bit confused on these parts as well, they really helped!
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