# f325 question

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#1
on this paper
http://www.ocr.org.uk/Images/65869-q...d-elements.pdf

how is the answer for q7 d ii obtained?

here is the markscheme
http://www.ocr.org.uk/Images/60302-m...ments-june.pdf
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#2

that should get enough attention.
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6 years ago
#3
You know the over all equation, try to work out two half equations one for each product. Trial and error.
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6 years ago
#4
(Original post by nmjasdk)
on this paper
http://www.ocr.org.uk/Images/65869-q...d-elements.pdf

how is the answer for q7 d ii obtained?

here is the markscheme
http://www.ocr.org.uk/Images/60302-m...ments-june.pdf
You are told that the electrolyte is aqueous hydroxide ions.
You are told the products.
There must be an oxidation (loss of electrons) and a reduction (addition of electrons).
The electrons must balance to give the overall equation.

2H2O --> 2H2 + O2

4OH- - 4e -> 2H2O + O2
4H2O + 4e --> 4OH- + 2H2

the second half equation can be divided through by 2 to give
2H2O + 2e --> 2OH- + H2

although glancing at the markscheme, the doubled version is allowed ...
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6 years ago
#5
(Original post by charco)
You are told that the electrolyte is aqueous hydroxide ions.
You are told the products.
There must be an oxidation (loss of electrons) and a reduction (addition of electrons).
The electrons must balance to give the overall equation.

2H2O --> 2H2 + O2

4OH- - 4e -> 2H2O + O2
4H2O + 4e --> 4OH- + 2H2

the second half equation can be divided through by 2 to give
2H2O + 2e --> 2OH- + H2

although glancing at the markscheme, the doubled version is allowed ...
This.

Another tip and good practice for redox questions is to start by writing the half equation for oxidation above the half equation for reduction. In this case, the half equation for oxidation gives clues as to the reduction step.
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