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Would a high voltage ( of the order of a few score kV ) drive electrolysis?

Or, does it require a large current and low voltage? Alternatively, does electrolysis require both large voltage & large current?

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For any electrochemcial reaction, electrons are transfer from the oxidant to the reductant. By definition, any flow of charge per unit time period is an electric current.

However, in the case of electrolysis, the electron transfer is not spontaneous. An external energy source is required for the reaction to take place. The energy (work) provided per unit charge is called the voltage.

For any electrolysis reaction to occur therefore requires a supply of electrical energy which means both a voltage and electric current are needed.

Specifically, the amount of work (energy) resulting from a voltage, $V$ and a current, $I$ applied for a time $t$ is given by:

Work produced by electrical energy, $W = V.I.t$

For a given electrolytic reaction to proceed, the required minimum voltage which must be applied is determined from the Gibb's function,

$E^{0}_{cell}=\frac{-\triangle G^{0}}{nF}$

Where $E^{0}_{cell}$ is the minimum voltage needed for the electrolysis reaction to occur, $\triangle G^{0}$ is the change in Gibb's free energy under standard conditions, $n$ is the number of electrons transferred and $F$ is Faraday's constant (96,485 coulombs per mole).

For example, if pure water is placed in an 'electrolytic cell' with two non-reactive electrodes (eg: platinum), electrons forced into the 'cell' by an electrical source (eg: battery) will react with water molecules at the cathode forcing them to lyze (split) into hydrogen ions and hydroxide ions.

$H_{2}O(aq)\xrightarrow{elect}H^{+}(aq)+OH^{-}(aq)$

At the surface of the 'anode', hydroxide ion will be oxidised (dontate electrons) to form oxygen gas.

$4OH^{-}(aq)\rightarrow 2H_{2}O+O_{2}+4e^{-}$ (anode half-cell reaction)

Meanwhile, at the surface of the 'cathode', hydrogen ions will accept electrons to form hydrogen gas.

$2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)$ (cathode half-cell reaction)

The overall cell reaction is therefore:

$2H_{2}O(aq)\rightarrow2H_{2}(g)+O_{2}(g)$ ; $\triangle G^{0}=237.2kJ/mol$

The positive Gibb's function for this reaction indicates that the reaction will not occur sponteneously but requires an external energy source (eg: electrical energy), which is reflected in the negative cell potential.

The minimum (theoretical) voltage required is:

$E^{0}_{cell}=\frac{-\triangle G^{0}}{nF}=\frac{-237.2 \times 10^{3}}{2\times 96485}=-1.23V$

However, in practice, a slightly larger voltage of 1.48V is required since the enthalpy (heating) of the products results in slightly lower efficiency which is manifested as an overpotential of about 0.25V.

Once this critical voltage level is exceeded, the electrolysis reaction proceeds at a rate determined largely by the the current, since the current represents the rate at which charge is delivered to the system. Basically, the higher the current, the more molecules will react (electrolyze) and the more products (hydrogen gas) will form per unit time.

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Current is, by definition, a flow of charged particles, so it doesn't drive anything. Electrical potential is a store of energy (you can think of a voltage source as an ideal "battery" of sorts).

One volt can produce 1 A of current through a 1 ohm resistor. One ampere of current is equal to a coulomb per second. One coulomb is 6.24150965(16)×10^18 electrons. Were you to use enough voltage that oxidation at the anode released enough electrons to produce a massive current, you'd likely deplete your anode material before you ever achieved it.

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