Why does the current of a galvanic cell decrease with time?

1 Answer
May 17, 2018

The reaction is approaching equilibrium in that time. At equilibrium, the cell potential is zero, and neither reaction direction dominates anymore (the rates are equal and nonzero).


We know that Q is the reaction quotient, and Q = Q(t). We also know that

lim_(t->oo) Q(t) = K,

the equilibrium constant. Consider that the Nernst equation expresses the cell potential at non-equilibrium conditions, i.e. when the cell still has free energy available to do electrical work:

E_(cell) = E_(cell)^@ - (RT)/(nF)lnQ

or

E_(cell)^@ - E_(cell) = (RT)/(nF)lnQ

This must be true at equilibrium:

cancel(E_(cell))^(0) = E_(cell)^@ - (RT)/(nF)lncancel(Q)^(K)

and

cancel(DeltaG)^(0) = -nFcancel(E_(cell))^(0)

Furthermore,

  • Spontaneous reactions have DeltaG < 0 and E_(cell) > 0.
  • Nonspontaneous reactions have DeltaG > 0 and E_(cell) < 0.

Over time, Q -> K, whether Q > K or Q < K.

  • If Q < K, then the forward reaction is spontaneous and Q increases towards K so that E_(cell) decreases until E_(cell) = 0 at equilibrium. It also follows that DeltaG increases until DeltaG = 0 at equilibrium.
  • If Q > K, then the reverse reaction is spontaneous and Q decreases towards K so that E_(cell) increases until E_(cell) = 0 at equilibrium. It also follows that DeltaG decreases until DeltaG = 0 at equilibrium.

And when E_(cell) = 0, there is no potential for the electrons to flow any more, and we have a cell at equilibrium.