100 cm³ of 2 M sulphuric (VI) acid was electrolyzed using the set-up represented by the diagram below

100 cm³ of 2 M sulphuric (VI) acid was electrolyzed using the set-up represented by the diagram below

i) Write an equation for the reaction that produces gas L.
ii) Describe how gas K can be identified
iii) Explain the difference in:
I. The volume of the gases produced at the electrodes.
II. Brightness of the bulb if 100 cm³ of 2 M ethanoic acid was used in place of sulphuric (VI) acid.

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i) $4OH_{(aq)}^- \rightarrow O_{2(g)} + 2H_2O_{(l)} + 4e^-$

ii) Insert a burning splint in gas K. The gas burns with a pop sound

iii) I. For every one molecule of oxygen formed two molecules of hydrogen are produced hence the

difference in volume.

II. The bulb is brighter with sulphuric (VI) acid. SUlphuric (VI) acid is a strong acid hence its degree
of ionization is higher while ethanoic acid is a weak acid hence has a few ions since it ionizes
partially.

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johnmulu answered the question on April 27, 2017 at 07:49

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