Test 2 Q8

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Tim Foster 2A
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Test 2 Q8

Postby Tim Foster 2A » Mon Mar 05, 2018 3:04 pm

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?

Naveed Zaman 1C
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Re: Test 2 Q8

Postby Naveed Zaman 1C » Mon Mar 05, 2018 3:59 pm

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).

Michelle Chernyak 1J
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Joined: Fri Sep 29, 2017 7:06 am

Re: Test 2 Q8

Postby Michelle Chernyak 1J » Mon Mar 05, 2018 4:31 pm

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-]

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