# 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# and#E_(cell) > 0# .*Nonspontaneous*reactions have#DeltaG > 0# and#E_(cell) < 0# .

Over time,

- 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