The diagram given below shows the electrolysis of acidulated water. With reference to the diagram, answer the following questions:
(a) Write an equation for the reaction that takes place at A.
(b) Give a test to confirm the product formed at A.
(c) Which gas is liberated at B?
(d) Why is water acidified for electrolysis?
(e) What is the ratio by volume of the gases liberated at A and B?
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(a) A is the anode (positive electrode) where oxygen gas is liberated.
4OH- - 4e- ⟶ 2H2O + O2
(b) The gas at A is oxygen. It rekindles a glowing splinter brought near it.
(c) Hydrogen gas is liberated at B.
(d) Pure water is a very poor conductor of electricity as it has very few ions. Water is acidified by adding a little dilute sulphuric acid to provide ions, which makes it a good conductor so that electrolysis can take place.
(e) The ratio by volume of gases at A (oxygen) and B (hydrogen) is 1 : 2.