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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
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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.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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