Question

In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change? 2HClO4 (aq) + CaCO3 (s) → Ca(ClO4)2 (aq) + H2O (l) + CO2 (g) CaO (s) + H2O (l) → Ca(OH)2 (s) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g) SO2 (g) + H2O (l) → H2SO3 (aq)

Answer 1

Answer: The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Explanation:

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

• Reaction 1:  $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

• Reaction 2:  $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

• Reaction 3:  $HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

• Reaction 4:  $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

• Reaction 5:  $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$