Question

The discovery of the fact that oxygen and hydrogen combine to form water was a major step in the development of science. Who made this discovery?
a) Carl Wilhelm Scheele
b) Isaac Newton
c) Joseph Priestley
d) Henry Cavendish

Answer 1

Hint:Formation of water takes place by the combination reaction of hydrogen and oxygen. In this process two moles of hydrogen is needed with one mole of oxygen. According to the law of definite proportion, the ratio always remains constant.

Complete answer:
Scientist Henry Cavendish discovered the composition of water.
He was doing an experiment by using oxygen and hydrogen as reactants, and mixed those elements together, thinking it will create an explosion by oxyhydrogen effect.
Water has a $s{p^3}$ hybridised structure. It acquires bent shape as the two lone pair on oxygen which pushes the $O - H$ bond in downward direction by disturbing its linear geometry.
But the discovery of formula of water i.e. ${H_2}O$ Was made by Amedeo Avogadro.
As we know water exists in three states-solid as ice, liquid- as water and gas as steam.
By the formula of water we can calculate the molar mass, as the molar mass of hydrogen is $1gm$ and the molar mass of oxygen is $16gm$
$\begin{gathered}
  \therefore {H_2}O = 2 \times 1gm + 16gm \\
   \Rightarrow {H_2}O = 18gm \\
\end{gathered} $
Also we can relate the value as $1$ mole of ${H_2}O$ has $18gm$ which contains $6.022 \times {10^{23}}$ number of molecules of water.
The reaction of formation of water can be written as the following-
$2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l)$
Electrolysis of water can be explained as the process of decomposition of water into oxygen and hydrogen gas at respective electrodes.
In this method different reactions are observed at the two electrodes namely anode and cathode.
In the electrolytic process the decomposition reaction of water in acidic medium can be written as-
$\begin{gathered}
  Cathode:2{H^ + }(aq) + 2{e^ - } \to {H_2}(g) \\
  Anode:2{H_2}O(l) \to {O_2}(g) + 4{H^ + }(aq) + 4{e^ - } \\
\end{gathered} $
Now balancing the equation at cathode-
$Cathode:4{H^ + }(aq) + 4{e^ - } \to 2{H_2}(g)$
Thus the net reaction after adding the two half-cell reaction, the full-cell reaction in the electrolysis can be represented by-
$Net:2{H_2}O(l) \to 2{H_2}(g) + {O_2}(g)$

Hence the correct option is (d).

Note:
After electrolysis hydrogen undergoes reduction and the gas evolved at cathode as the sign convention of this cathode is negative but oxygen undergoes oxidation and moves to anode as the sign convention of this electrode is negative.