6L.7a

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Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

6L.7a

Postby Lindsey Chheng 1E » Sun Mar 01, 2020 3:46 pm

How do you get the half reactions for AgBr(s) ⇌ Ag+(aq) + Br-(aq)
The anode is supposed to be Ag(s) + Br- ⟶ AgBr(s) + e- and the cathode is supposed to be Ag+(aq) + e- ⟶ Ag(s). Am I supposed to get these on my own or am I supposed to just look in appendix 2a in the back of the book?

Connor Chappell 2B
Posts: 59
Joined: Wed Feb 20, 2019 12:16 am

Re: 6L.7a

Postby Connor Chappell 2B » Sun Mar 01, 2020 3:51 pm

You can derive these half-reactions by the change in oxidation states of both of the species in the reaction. You notice that in the reaction given, Br is neither oxidized nor reduced, and Ag is being both oxidized and reduced in the reaction. Therefore, Ag is both the oxidizing and reducing agents, and the half-reactions can be derived from this recognition.


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