Question

Electrolysis of acidified water using platinum electrodes gives___________.
A. Oxygen and Hydrogen peroxide
B. Hydrogen and oxygen
C. Hydrogen peroxide
D. Hydrogen and hydrogen peroxide

Answer 1

Hint: Water is a poor conductor of electricity but it has some \[{H^ + }\]hydrogen and \[O{H^ - }\] hydroxide ions which formed after dissociation of water molecules. So we have added some acids like dilute Sulfuric acid to carry out electrolysis.

Complete solution:
Let us understand how electrolysis of acidified water is carried out.
As we know ionization of water is very weak and produces only minute concentration of hydrogen ions and hydroxide ions, so when sulfuric acid is added which produces hydrogen ions \[{H^ + }\] and sulfate ions $S{O_4}^{2 - }$ from the acids, makes water good conductor of electricity.
Now we will learn what reactions are taking place at anode and cathode.
1. At cathode: Here reduction takes place i.e. electron gain. The hydrogen ions \[{H^ + }\]are produced and will attach to cathode and will discharge as ${H_2}$ gas.
$2{H^ + } + 2{e^ - } \rightleftarrows {H_2}(g)$
2. At anode: Here oxidation takes place i.e. electron loss. The negative sulfate $S{O_4}^{2 - }$ and \[O{H^ - }\] hydroxide ions are attached to the positive electrode.
Here both ions compete as an anode to oxidize $O{H^ - }$ but win and oxidize.
$4O{H^ - } \to 2{H_2}O + {O_2} + 4{e^ - }$

Overall reaction:
$2{H_2}0 \rightleftarrows 2{H_2}(g) + {O_2}(g)$
The ratio of two gases evolved is in the ratio of 2:1.This will help us to solve numerical problems.

So our correct option is B. Hydrogen and oxygen

Additional information:
Now let us learn how we will test the gases which evolved. Hydrogen gas gives a pop sound and oxygen gas rekindles a glowing splint.

Note:
Remember the reactions thoroughly and the main thing to learn is that at anode only hydroxide ions oxidize, not sulfate ions. Ratio of gases produced is 2:1.