Why does the current of a galvanic cell decrease with time?
1 Answer
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
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 andE_(cell) > 0 . - Nonspontaneous reactions have
DeltaG > 0 andE_(cell) < 0 .
Over time,
- If
Q < K , then the forward reaction is spontaneous andQ increases towardsK so thatE_(cell) decreases untilE_(cell) = 0 at equilibrium. It also follows thatDeltaG increases untilDeltaG = 0 at equilibrium. - If
Q > K , then the reverse reaction is spontaneous andQ decreases towardsK so thatE_(cell) increases untilE_(cell) = 0 at equilibrium. It also follows thatDeltaG decreases untilDeltaG = 0 at equilibrium.
And when