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Redox Reactions — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryRedox Reactions5 Marks
Question
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
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Answer

  1. C is a reducing agent while $\mathrm{O}_2$ is an oxidising agent. If excess of carbon is burnt in a limited supply of $\mathrm{O}_2, \mathrm{CO}$ is formed in which the oxidation state of C is +2 . If, however, excess of $\mathrm{O}_2$ is used, the initially formed CO gets oxidised to $\mathrm{CO}_2$ in which oxidation state of C is +4 .
$2\text{C(s)}+\text{O}_2(\text{g})\rightarrow\ \stackrel{{+2}}{\ \ 2 \text{CO(g)}};\ \ \text{C(s)}+\text{O}_2(\text{g})\rightarrow\ \stackrel{{+4}}{\ \ \text{CO}_{2}}(\text{g})\$\text{Excess})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Excess})$
  1. $P_4$ is a reducing agent while $Cl_2$ is an oxidising agent. When excess of $P_4$ is used, $PCl_3$ is formed in which the oxidation state of P is +3. If, however, excess of $Cl_2$ is used, the initially formed $PCl_3$ reacts further to form $PCl_5$ in which the oxidation state of P is +5.
$\text{P}_4(\text{s})+6\text{Cl}_2(\text{g})\rightarrow\stackrel{{+3}}{4 \text{PCl}_3};\ \ \ \text{P}_4(\text{s})+10\text{Cl}_2\rightarrow\ \stackrel{{+5}}{4 \text{PCl}_5}\\ (\text{Excess})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Excess})$
  1. Na is a reducing agent while $O_2$ is an oxidising agent. When excess of Na is used, sodium oxide is formed in which the oxidation state of O is -2. If, however, excess of $O_2$ is used, $Na_2O_2$ is formed in which the oxidation state of O is -1 which is higher than -2.
$4\text{Na(s)}+\text{O}_2(\text{g})\rightarrow\stackrel{{-2}}{ \text{Na}_2\text{O(s)}};\ \ 2\text{Na} (\text{s}) +2\text{O}_2(\text{g})\rightarrow\ \stackrel{{-1}}{ \text{Na}_2\text{O}_2(\text{s})}\\ (\text{Excess})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Excess})$

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