HW 14.1
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Olivia_Chen_3E
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HW 14.1
In the homework for 14.1, the answer in the back of the book for part (b) the reduction half reaction, involves 14H-. I was just wondering what exactly does H- mean and for the final addition of the two half reactions for part (d) how this cancels with the oxidation reaction to give us 8H+.
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Britney Pheng 1L
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Re: HW 14.1
Hi Olivia!
Pretty sure that's a typo in the book and it's supposed to be 14H+.
In regards to how you get 8H+ in the end, you have to multiply part b) by 3 to get the same amount of electrons for both half-reactions before you can add them. This means you will have 6H+ on the products side, and 14H+ on the reactants side. You can then subtract 6H+ from both sides to leave you with 8H+ on the reactants side.
Hope that helps!
Pretty sure that's a typo in the book and it's supposed to be 14H+.
In regards to how you get 8H+ in the end, you have to multiply part b) by 3 to get the same amount of electrons for both half-reactions before you can add them. This means you will have 6H+ on the products side, and 14H+ on the reactants side. You can then subtract 6H+ from both sides to leave you with 8H+ on the reactants side.
Hope that helps!
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