I should show the kids how electrolysis could create $\ce{H2}$. We used a drinking glass of water mixed with salt. Then we put an 1.5 volt AA battery in the glass and the electrolysis was visible after about 30 sec.

We then added red cabbage juice which after some hours turned from violet to blue/green. So it must obviously had become base.

Is this the reaction?:

$\ce{H2O + NaCl -> H2 + O2 + NaOH + Cl2}$

Or could the nickel in the battery have an impact on the result?

Salt often contains $\ce{KCl}$ but that might just give $\ce{KOH}$ which also is a base.

I guess the amount of chlorine gas ($\ce{Cl2}$) is so small that it is not a health risk.


1 Answer 1


Electrolysis reaction

The electrolysis of sodium chloride solution proceeds by the following reaction:$$\ce{2NaCl (aq) + 2H2O (l) -> H2 (g) + Cl2 (g) + 2NaOH (aq)}$$

The absence of a membrane cell

Since you carry out the reaction without a membrane cell, there is no separation of liberating chlorine from forming sodium hydroxide solution. The absence of the membrane cell causes chlorine disproportionation in alkali solution. So, the following reaction occurs under standard conditions (20-25°C):$$\ce{2NaOH (aq) + Cl2 (g) -> NaOCl (aq) + NaCl (aq) + H2O (l)}$$ Heating of the resulting solution leads to further oxidation of $\ce{[OCl]-}$ to $\ce{[ClO3]-}$:$$\ce{3[OCl]- (aq) -> [ClO3]- (aq) + 2Cl- (aq)}$$

Nickel impact

It is needed more information about the battery itself to determine if nickel from it can have an impact on the electrolysis reaction.


  1. R.J.D. Tilley (2004). Understanding Solids: The Science of Materials: p. 281.
  2. C.E. Housecroft, A.G. Sharpe. Inorganic Chemistry (4th ed.): "The group 17 elements", p. 615.
  3. N.N. Greenwood, A. Earnshaw. Chemistry of the Elements (2nd ed.): "Industrial Production and Uses of Chlorine", p. 798.
  • $\begingroup$ So no O2 will be produced? $\endgroup$
    – hschou
    Jul 26, 2022 at 8:11
  • 1
    $\begingroup$ @hschou, it depends on a concentration of sodium chloride in the solution. If $\ce{NaCl}$ solution is concentrated then there will be mostly liberation of chlorine. In case of low concentration of $\ce{NaCl}$ in the solution there will be liberation of chlorine and oxygen simultaneously (since the electrolysis of water on the anode). Compare two electrode potentials:$$\ce{2Cl- (aq) -> Cl2 (g) + 2e-, E^\circ= -1.36 V}$$$$\ce{2H2O (aq) -> O2 (g) + 4H+ (aq) + 4e-, E^\circ= -1.23 V}$$References: W.M. Haynes. CRC Handbook of Chemistry and Physics (95th ed.): Electrochemical Series, 5-80. $\endgroup$
    – Exeplone
    Jul 26, 2022 at 9:25
  • 1
    $\begingroup$ See also: youtube.com/watch?v=1nAe_T5mO2o $\endgroup$
    – Exeplone
    Jul 26, 2022 at 9:42

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