At school we've learned about calculating the voltage needed to perform an electrolysis. The current is how many electrons pass per time. The formula is given $$I = \frac UR$$

This means that given an electrolysis with $U=\pu{4.0 V}$ we get $$I = \frac{\pu{4.0 V}}{R}$$

If I am correct, what would decide how fast the electrolysis will go, is the resistance. Is the resistance fixed, or can I do anything to lower it? How would increasing the voltage affect the electrolysis? Would it just make it go faster (because $I$ is increased)?


2 Answers 2


Voltage determines the kind of chemistry at each electrode. Gold is purified by plating the anode to the cathode at minimum voltage for the half-reaction, so impurities either do not dissolve and do not plate out. Copper electrorefining electrode mud contains valuable trace elements in concentrated from.

Current determines the amount of chemistry, ~96,500 coulombs/mole of electrons.

Cell internal resistance converts input energy, power multipleid by time, $\ce{I^2Rt}$, into useless heat. You then want small electrode spacing and good electrolyte stirring. You then need enough voltage to do the chemistry, plus more to compensate for resistance losses, but not so much voltage that undesirable chemistries are enabled. Does cell resistance increase (ohmic) or decrease with temperature?

  • $\begingroup$ Thank you for a good answer! So you say that the voltage should be as high as possible without enabling undesired reactions to increase the current? Additionally have a small distance between the electrodes to reduce the resistance? $\endgroup$ Commented Feb 14, 2014 at 21:56
  • $\begingroup$ How many ampere should I get with 5 cm spacing, $3 V$ and a saline NaCl-solution in water? Is it safe to connect a car battery charger? $\endgroup$ Commented Feb 14, 2014 at 21:58
  • $\begingroup$ "I2Rt, into useless heat" of only minor importance: While all waste heat is lost energy in that it does not contribute to actual electrolysis, a hot electrolyte is more conductive then a cold one, thus retained heat will increases the rate of electrolysis. Since rate of electrolysis is effected by surface area, strength of electrolyte and heat, then to a degree, even waste heat can be an important contributing factor of a production unit. $\endgroup$ Commented Jun 23, 2015 at 18:04

Regarding your question: "Is the resistance fixed, or can I do anything to lower it?"

The energy used in electrolysis produces both product and waste heat. Naturally, you want to maintain sufficient production while limiting wasted energy.

A few important formulas:

  • Amps = volts / resistance
  • Watts = amps * volts
  • Joules = watts * seconds.

Things that effect resistance, and their effects on product versus waste heat and rate of production: Adjusting each parameter individually:

  • Voltage: Per unit of energy supplied, in general, a lower voltage produces more product and less waste heat from electrolysis. Below a certain threshold voltage no electrolysis will occur. Although less power is consumed to produce the same product at a low voltage at a higher efficiency, the process proceeds slower because with a lower voltage, there is less current and power is consumed more slowly. Alternativly one can increase the voltage to drive more current and produce their product faster, but this happens at an expence to efficency.

  • Electrolyte type and concentration: Different electrolytes have different capacities to conduct and Its concentration above or below an ideal will negatively effect the electrolyte's capactiy to conduct electricity.

  • Temperature: Increasing the temperature of electrolyte lowers the resistance of an electrolyte. Additionally, some of the energy used to break up water's molecular bonds is provided by the thermal energy of molecules bouncing off each other. But, the effect really only becomes appreciable with high temperature electrolysis, like say 800c.

  • Surface Area: Surface area and current are proportional. All other parameters being equale, if you double the surface area of your electrodes, you effectivly half the resistance and double the current.

  • Spacing: Decreasing the spacing between electrodes decreases resistance.


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