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I am having a galvanic cell with the following half cell equations:

Anode--> 2Al(s) + 6OH --> 2Al(OH)3 + 6e-

Cathode--> 6H+ + 6e- --> 3H2(g)

I connected my cell to a 47 ohm resistor and measured the voltage across the load, which was 0.07V. I used ohms law to calculate current:

$$\frac{0.07V}{47Ω}$$ = 1.5mA. I wanted to know why my current is so low even though i am producing 6 electrons.

My anode electrolyte is 0.02 Molar aluminum sulfate and cathode electrolyte is 0.00026 Molar acetic acid

The open-circuit voltage is 1.34 volts

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    \$\begingroup\$ maybe you need to produce 6 electrons more often. \$\endgroup\$ – Brian Drummond Mar 12 '16 at 23:12
  • \$\begingroup\$ 47 Ω might be a bit on the low side for a test load. What's the open-circuit voltage? Try measuring the voltage with a 470 Ω and then a 4.7 kΩ load as well as open-circuit and report back (in your question rather than in the comments). \$\endgroup\$ – Transistor Mar 12 '16 at 23:17
  • \$\begingroup\$ @transistor why should I go with a higher resistor if the lower resistor value produced a low current \$\endgroup\$ – user510 Mar 12 '16 at 23:24
  • \$\begingroup\$ and i added the open-circuit voltage \$\endgroup\$ – user510 Mar 12 '16 at 23:26
  • \$\begingroup\$ See my answer below. If anything is not clear then please ask and someone will answer if I can't. \$\endgroup\$ – Transistor Mar 13 '16 at 0:09
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My chemistry isn't so good but I expected that you would be hoping for a 1 to 2 volt cell. You don't give any details of the size of your cell but I imagine that it's small or that the surface area of the electrodes may be small and that's limiting the maximum number of electrons it can produce.

All cells have internal resistance which causes the voltage to droop in proportion to the current drawn. We can model a cell (or battery) as a "perfect" voltage source of potential equal to the open-circuit voltage with a series-equivalent-resistor. By measuring the open-circuit voltage and making a test a known load and we can calculate the equivalent internal series resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1 and 2. Open-circuit and loaded voltage measurements.

  • For the open-circuit measurement we use a voltmeter with a high input impedance so that negligible current is drawn from the cell. You got 1.34 V.
  • We take a second reading with the 47 Ω load. You got 0.07 V.
  • The voltage drop across the internal resistance is then 1.34 - 0.07 = 1.27 V.
  • Let's double-check your current reading by calculating the current through RL, 47 Ω, using Ohm's Law. \$ I = \frac {V}{R} = \frac {0.07}{47} = 0.00149~A = 1.49~mA \$ which is very close to your reading. Excellent.
  • We can now calculate \$ R_S = \frac{V}{I} = \frac {1.27}{0.00149} = 852~Ω \$.

With that calculation we can see that if we loaded the cell with an external 852 Ω resistor we would only get half voltage, 0.67 V, at the terminals. If you use a lower resistor (bigger load) the voltage will drop further.

Maximum power

If you leave your cell open-circuit you can get maximum voltage out of it. This isn't much use as the current is zero and since \$ P = V \cdot I \$ the power out is zero.

If you short circuit the battery you can get maximum current out of it but with zero volts across the short you get zero power again. (All the power is dissipated in the internal resistance.

enter image description here

Figure 3. Power transfer graph. Source: Wikipedia Commons. (See link below.)

The maximum power transfer theorem shows that the most power you can get from the cell is when \$ R_L = R_S \$.


Other info

The ampere is defined as "a unit of electric current equal to a flow of one coulomb per second".

A coulomb is equivalent to the charge of approximately \$ 6.242 × 10^{18} \$ electrons.

You're getting 1.5 mA so that's \$ 6.242 × 10^{18} × 0.0015 = 9.36 × 10^{15} \$ electrons per second. Congratulations!

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  • \$\begingroup\$ ya I used aluminum can as my anode electrode and i used a 1 by 5 inch platinum sheet. How will the resistivity of these electrodes and electrolyte mentioned in the OP effect current: will it be significant or negligible? \$\endgroup\$ – user510 Mar 13 '16 at 0:49
  • \$\begingroup\$ If the temperature of the cell is near 2˚C, how does this affect conductivity? \$\endgroup\$ – user510 Mar 13 '16 at 1:18
  • \$\begingroup\$ I would imagine that the resistivity of the electrodes is very low compared with the calculated 852 Ω internal resistance. You might find that the oxide layer on the aluminium affects performance. Since it oxidises instantly in air I don't know how to avoid that. Charge mobility in the electrolyte will affect current. I have no idea about cell temperature. \$\endgroup\$ – Transistor Mar 13 '16 at 9:27
  • \$\begingroup\$ yeah online they say that the lower the temperature, the more the resistance of the flow of ions/charge \$\endgroup\$ – user510 Mar 13 '16 at 15:19
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The open circuit voltage \$V_{Cell}\$ is based on the difference of the two electrode potentials. By connecting the electrodes over a load (the resistor \$R_L\$) the galvanic cell has to sustain a current of \$I=\frac{V_{Cell}}{R_L}= \frac{1.34V}{47\Omega}= 28.5 mA\$ according to Ohm's law and the values given in the question.

To do so the reaction has to happen frequently enough to let enough charge per time unit flow (= current). This depends on the surface area of the electrode, electrolyte concentration in the cells and temperature amongst others.

The number of electrons flowing per second to sustain such current is orders of magnitude higher than the 6 electrons per reaction. Given that one Ampere is the flow rate of \$6.24\cdot10^{18}\$ electrons per second (Wiki), the cell has to supply a flow rate of \$1.78\cdot10^{17}\$ electrons per second.

If you model your cell from a electrical perspective one would assume a voltage source with a high(in this case) internal series resistance.

This answer is focused on the main issue and skips modeling details found in (real=non-ideal) batteries or galvanic cells.

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