Question

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

Answer 1

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).

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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.