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In an experiment, I set up a cell with lead nitrate (w/ lead electrode) and zinc sulfate (w/ zinc electrode), with a salt bridge containing potassium nitrate.

I observed that by increasing the number of salt bridges, the voltage measured increased. However, given what I know about electrochemical cells, I can't work out why. I initially thought it would have no effect on the voltage/potential difference.

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    $\begingroup$ I suspect this has to do with the kinetics of ion transfer between the two cells through the aqueous salt bridge. If the bridge is too thin, then maybe it could acting as a choke. Was the initial bridge very thin? Did each additional bridge cause the same increase in voltage, or did it asymptotically approach a value? $\endgroup$ – Nicolau Saker Neto Jun 7 '14 at 14:19
  • $\begingroup$ @NicolauSakerNeto: The relationship was linear, i.e. voltage as function of salt bridges N was of the form $V(N)=aN+b$. The salt bridge I used was filter paper soaked in the salt. $\endgroup$ – user1997744 Jun 7 '14 at 15:01
  • $\begingroup$ I think your bridge was sub optimal. What voltage was you able to reach? $\endgroup$ – Jori Jun 8 '14 at 10:58
  • $\begingroup$ @Jori: With 5 layers of filter paper, around 350 mV. Can you explain why increasing the number of alt bridges increased the voltage? $\endgroup$ – user1997744 Jun 8 '14 at 11:00
  • $\begingroup$ 350 mV?! You should be able to get around 630 mV. Like Nicolau Saker Neto said, it seems that your bridge isn't working properly. I don't know about the ability of Potassium nitrate to conduct electron transfer, perhaps that could be the problem. Try soaking your bridge in Potassium iodide (table salt can also be used, albeit less optimal). What martial is your bridge made of? $\endgroup$ – Jori Jun 8 '14 at 11:35
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The oxidation process either produces positive ions or removes negative ions from the solution at the anode (or it may change one ion to a more positive one), and the reduction process either removes positive ions or produces negative ions in the solution at the cathode. This produces electrically charged solutions, and very quickly stops the process before a measureable number of electrons are transferred. There must be a path for the ions to move between the two solutions in order for electrons to flow continuously through the wire. This produces an "ion current" within the battery with cations (positively - charged ions) moving from anode to cathode, and anions (negatively - charged ions) moving from the cathode toward the anode.

enter image description here

Now, more the no. of salt bridges, more will be the ion current. And from the standard ohms law equation, $$V=IR$$ , V being directly proportional to I (Current), will increase if the ion current increases.

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    $\begingroup$ Only up to a point, you will never beat the equilibrium voltage. $\endgroup$ – Kevin Kostlan May 9 '15 at 16:46
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The voltage you measure between the terminals of a voltaic cell will depend on two factors:

  1. The intrinsic maximum voltage $(V_\mathrm{max} = E_\mathrm{cell})$ that the cell could produce, depending on the $E^o_{red}$ of each half cell, the ion concentrations and the temperature. This is calculated from the Nernst equation:

    $$E_\mathrm{cell} = E^⦵_\mathrm{cell} - \frac{RT}{zF}\ln{Q_r}$$

  2. Resistive terms that come from within the cell, including overpotentials at the electrodes and other factors that influence the internal resistance of the cell.

Overall, the observed voltage $(V_L)$ measured at the terminals of the cell will be lower than the theoretical $V_\mathrm{max}$ because the internal resistance of the cell involves a voltage drop, in the same way that you would see an external voltage drop across a resistor in series in the circuit.

$V_L$ can be calculated in the following way:

$$V_L = V_\mathrm{max} - V_I$$

where $V_I$ is the internal voltage drop from the internal resistance of the cell. This drop in voltage depends on the internal resistance of the cell ($R_I$) and the total current $(I)$ running through the cell (and the circuit, see farside.ph.utexas.edu):

$$V_I = I \cdot R_I$$

Your question seems to revolve around how the salt bridge affects the internal resistance of the cell. It would make sense that the more salt bridges you add, the less internal resistance would result, so the smaller the voltage drop and the larger the measured voltage from the cell.

It is possible to think of the salt bridge, which involves the flow of ions, as being analogous to a wire, which involves the flow of electrons. The resistance of a wire $(R)$ is dependent on it cross-sectional area (see http://hyperphysics.phy-astr.gsu.edu):

$$R = \frac{ρL}{A}$$

where $ρ$ is resistivity; $L$ is length; $A$ is cross sectional area.

So it follows that the same may be true for a salt bridge. By adding more salt bridges, you are increasing the cross-sectional area, decreasing the internal resistance and increasing the observed voltage, which would approach the theoretical maximum voltage as the internal resistance approaches zero.

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