Electrolysis of molten lithium hydroxide at $\pu{750 K}$ to form lithium metal

$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.$$

Electrolysis will only proceed at an appreciable rate when the applied potential exceeds the electrochemical cell potential by $\pu{0.60 V}.$

(c) Calculate the minimum potential that should be applied.

$$ΔG^\circ = -nFE^\circ_\mathrm{cell}$$

$$E^\circ_\mathrm{cell} = \frac{ΔG^\circ}{nF} = \pu{-2.60 V}$$

Why should the difference of $\pu{0.60 V}$ be added onto the absolute value of $E^\circ_\mathrm{cell}~(+2.60)$ and not $-2.60$?

I understand $E_\mathrm{cell}$ is the measure of the potential difference between two half-cells in an electrochemical cell, but what is applied potential in this case? I thought electrochemical cells produce an electric potential difference (in this case $\pu{-2.60 V}),$ so why would they need an applied (positive) potential difference?


1 Answer 1


You are mixing two areas, electrode kinetics and electrode thermodynamics. If we have 3 apples and 2 oranges, this is not equal to 5 oranges or 5 apples but there are 5 fruits in total. Electrode thermodynamics tells you a "yes" or "no" answer in the sense that it is saying the reaction is spontaneous or not. Electrode kinetics tells you how fast the reaction will be.

When you wish to electrolyse something, it means that you want do a non-spontaneous reaction i.e., if you leave lithium in the presence of air and moisture, it will spontaneously form LiOH

$$\ce{4 LiOH <- 4 Li + 2 H2O + O2}.--(1)$$

Now, by forcing electric current through molten LiOH, you want to reverse this reaction and want the following to happen

$$\ce{4 LiOH -> 4 Li + 2 H2O + O2}.--(2)$$

The negative Ecell of -2.60 V only tells you that reaction (2) is not spontaneous at all. It has no other meaning whatsoever. That is, if you wait forever, you will not see any Li spontaneously forming out of LiOH.

Now as you read in the question, they are saying that if you apply exactly +2.60 V potential difference between two electrodes the rate of reaction is very slow. Remember negative sign was just for saying that the reaction is non-spontaneous. However if you apply +2.60+0.60 = 3.2 V of potential difference between the electrodes, the rate of formation of Li from molten LiOH will be quite fast. They don't say how fast but you will not have to wait for a long time.

So +2.60 V is the apple = electrode potential 0.6 V is the orange, it is called the overpotential in formal language.

You can only add them (not subtract) and get the total voltage required to start the electrolysis at an appreciable rate.

  • $\begingroup$ But what if I have a galvanic cell like fuel cell? What's the difference between cell potential and applied potential (or bias) in that setting? $\endgroup$
    – Shaun Han
    Commented Feb 18 at 13:46

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