Why Does a Battery Drain?

Discussion in 'Physics' started by roma, Apr 25, 2011.

  1. roma

    Thread Starter New Member

    Apr 20, 2011
    Why does an electrochemical cell drain?

    Imagine zinc at the anode and copper at the cathode. The zinc forms a Zn2+ electrolyte and the electrons travel through the circuit. They wind up at the copper cathode and are then transferred into the electrolyte that contains positive copper ions.
    A K2SO4 salt bridge connects the Zn2+ and Cu2+ electrolytes. The traditional explanation is that the salt bridge acts as a source of negative charges (sulfate anion) to the anode, and a source of positive charges (potassium cation) to the cathode. In this explanation, if it weren't for this replenishment, the anode would start to lose its negative charge as the electrons are sent through the wire to the cathode. The cathode would in turn start to lose its positive charge as the stream of electrons would neutralize its charge. The voltage between the anode and cathode would decrease and current would stop flowing. Even with the salt bridge, of course, the ion concentrations in the salt bridge eventually diminish and the battery eventually drains.

    The problem with this explanation is that it (seemingly) ignores the main requirement for a circuit to be complete in order for current to exist. If nothing besides the ion exchange described above occurs in the salt bridge, then how is the circuit closed? Without a closed circuit, there is voltage, but there is no current and so a lamp would not light up even for a millisecond if it were placed in the circuit.

    Another explanation that seemed to make sense to me was that the K+ ions in the salt bridge migrate toward the copper cathode side and accept the electrons that are coming into the electrolyte from the circuit through a porous membrane. These electrons are transferred to the SO42- side of the salt bridge. The SO42- ions have migrated to this side because they are attracted by the zinc cations in the anode elctrolyte. The electrons then pass through the SO42- ions, into the electrolyte and eventually back into the wire. The problem with this explanation is that it doesn't account for how the battery eventually drains.

    Please help :confused:
  2. designnut


    Apr 21, 2011
    zinc is exchanged wwith K in the solute. current causes the migration. Reveresing the current with external power can replate the zn, hence the rechargeable cell.
    Usually H2SO4 is used and there is a hydrogen exchange to provide the electron
  3. magnet18

    Senior Member

    Dec 22, 2010
    There is a complete circuit, but the current in the salt bridge is not electrons, rather charged ions which, as you stated, balance the charge out. This allows electrons to flow from the electrode with lesser reduction potential, to the one with the greater reduction potential (oxidation is loss, reduction is gain of electrons), this means that the electrode at the anode will lose mass as it's atoms go into solution as ions, and the cathode will gain mass as the concentration of it's solution goes down as the ions attach to the electrode and receive electrons. The electrons don't actually travel through the salt bridge, but rather the charged ions travel so that the overall charge in each solution remains neutral and the electrons will still flow.

    Also, the salt bridge is not the only reason the cell goes dead, if you have enough ions in the salt bridge the cell will stop operating long before they are used up.
    If you have a large enough salt bridge, the cell will go dead because the solution at the electrodes are getting more concentrated at the anode and less concentrated at the cathode. This can be calculated through the nernst equation, which is
    E_{cell} = E^o -\frac{.0591}{n} log(\frac{[RED]}{[OX]})

    (E_{cell} being the actual voltage, E^o being the standard voltage from a half reaction table(° always means standard), n being number of moles of electrons in the balanced equation, [RED] being the molarity of the anode solution, and [OX] being the molarity of the cathode solution(brackets always mean concentration))

    You can see through the math that as the anode concentration goes up and the cathode concentration goes down that the voltage of the cell will decrease.
    It is also quite possible that you simply run out of anode or that you run out of ions at the cathode, at which point the reaction can not continue.
    Last edited: Apr 25, 2011
  4. roma

    Thread Starter New Member

    Apr 20, 2011
    Thank you magnet18 for that detailed explanation! I think I'm all clear with it now.