6N.17


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905373636
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Joined: Thu Aug 01, 2019 12:15 am

6N.17

Postby 905373636 » Mon Mar 02, 2020 12:43 am

Why is there work done in this system, and how do we find Ecell? I keep calculating it to be 0V as the cathode and anode are reducing/oxidizing the same reaction in reverse to one another (?)

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

Re: 6N.17

Postby Hannah Lee 2F » Mon Mar 02, 2020 12:58 am

All components except the concentrations are the same, so we can infer that this is a concentration cell. Thus, E° = 0 (because under standard conditions, concentrations will be the same and so there will be no driving force for change).

wmax = deltaG = –nFE, so calculate E using the Nernst equation, with E° = 0:
E = E° - 0.05916 V / n (logQ) = -0.05916 V / n (logQ). For Q, plug in the concentrations given in the cell diagram. It might be easier to write out the half-rxns first to get a better idea of which concentrations to put as product and which to put as reactants, but generally, you always put the lower concentration in products (so in this case, Q = [anode] / [cathode]).

If E > 0, then deltaG < 0 so the cell rxn is spontaneous and can do work according to deltaG = -nFE.
If E < 0, then deltaG > 0 so the cell rxn is not spontaneous in that direction and cannot do work.

If E > 0, plug it into the formula for wmax to calculate the maximum amount of work the rxn can achieve per mole of Ag.


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