6L7 (a) HW

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Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

6L7 (a) HW

Postby Hannah Lee 2F » Thu Feb 27, 2020 4:27 pm

Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions:
(a) AgBr(s) --> <-- Ag1(aq) 1 Br2(aq), a solubility equilibrium

How can you tell which one is the anode and the cathode since the oxidation states are the same for both species?

Sebastian Lee 1L
Posts: 157
Joined: Fri Aug 09, 2019 12:15 am
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Re: 6L7 (a) HW

Postby Sebastian Lee 1L » Thu Feb 27, 2020 4:48 pm

In this case, the given reaction is the overall cell reaction and doesn't seem to have any change in oxidation numbers for reactants and products. However, the oxidation and reduction half-reactions will. This might not be the best way to determine the half-reactions, but I honestly just looked at the chart of reduction potentials (Appendix 2B) and saw that the reduction of AgBr yielded Ag (s) and Br- (aq). Based on this we can recognize that if AgBr is reduced, then the Ag (s) produced by the reaction would be oxidized into the Ag+ (aq) we have in the original reaction. However, when we look at the reduction potentials for the reactions, notice that the potential for the AgBr being reduced is only .07 but the potential for Ag+ being reduced is .80. Therefore, we know that to have a functional galvanic cell, the Ag+ must be reduced and the Br- (aq) must be oxidized into AgBr. This way we get a positive voltage so the reaction is favorable.

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