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Test 2 Q8

Posted: Mon Mar 05, 2018 3:04 pm
by Tim Foster 2A
when finding Q from the half reactions
Ag(s) + I-(aq) -> AgI(s) + e- E(anode)=-0.15V
AgCl(s) + e- -> Ag(s) + Cl- (aq) E(cathode) = 0.22V

How do you identify Cl- and I- as products or reactants?

Re: Test 2 Q8

Posted: Mon Mar 05, 2018 3:59 pm
by Naveed Zaman 1C
You have to balance the half reactions to get an overall redox reaction. In this case, the half reactions are already balanced, so the reaction is AgCl(s) + I-(aq) -> Cl-(aq) + AgI(s). So I- is a reactant and Cl- is a product, and Q = [Cl-]/[I-] (remember that solids are not included in this reaction).

Re: Test 2 Q8

Posted: Mon Mar 05, 2018 4:31 pm
by Michelle Chernyak 1J
When the two reactions are combined, Cl- is a product and I- is a reactant. Because we do not include solids or liquids when calculating the equilibrium quotient, Q= [Cl-]/[I-]