14.15 a?

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Adriana Rangel 1A
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14.15 a?

Postby Adriana Rangel 1A » Wed Mar 14, 2018 8:30 pm

I don't understand how to find the half reactions when given: AgBr(s) --> Ag+(aq) + Br-(aq), a solubility equilibrium.

I understand that at a cathode (reduction)occurs and anode (oxidation) occurs. I just don't understand the books explanation?
Does Ag and Br in AgBr(s) each have a charge of 0? Or do they have a charge of +1 & -1 respectively?
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Farah Ahmad 2A
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Re: 14.15 a?

Postby Farah Ahmad 2A » Wed Mar 14, 2018 8:48 pm

They have a charge of +1 and -1 respectively, so that the overall charge will be 0.
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Adriana Rangel 1A
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Re: 14.15 a?

Postby Adriana Rangel 1A » Wed Mar 14, 2018 9:39 pm

Ok and then what???
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