chem q (1 Viewer)

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for b wouldn't it b 11%, thx

 

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and also this please part a i dont know which oxidises or reduces

 

[Masaken]

and also this please part a i dont know which oxidises or reduces
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I forgot how to deal with concentrations with (w/v)% (will have to revise tomorrow), but for Q24, write out the full equation: Cu(s) + 2HCl(aq) --> CuCl2(aq) + H2(g). will assume that copper has the valence charge of 2+. Normally when determining which oxidises and which reduces you would look at the oxidation numbers, but I think since this equation is fairly simple you can achieve it by inspection: Cu(s) --> Cu2+ (aq) (ion that is part of the ionic compound CuCl2), and 2H+(aq) --> H2(g) on the other sides of the equation. 2Cl-(aq) is a spectator ion that's not considered as it remains an ion on both sides of the equation.

Oxidation is where electrons are lost, reduction is where electrons are gained. From that definition we can conclude Cu is getting oxidised (it loses 2 electrons to become the Cu2+ ion), and 2H+ is being reduced, as it gains electrons to become H2(g). You can find the half-equations from there.

 

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thx