Question

Iodate ion,$\text{IO}_{\text{3}}^{\text{-}}$ oxidises $\text{SO}_{\text{3}}^{\text{2-}}$ to $\text{SO}_{4}^{\text{2-}}$ in acidic medium. If $\text{100ml}$ sample of solution containing $\text{2}\text{.14g}$ of $\text{KI}{{\text{O}}_{\text{3}}}$ reacts with $\text{60ml}$ of $\text{0}\text{.5M}$ $\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$ solution, then final oxidation state of iodine is $\text{-x}$. The value of $\text{x}$ is:

Answer 1

In a redox reaction oxidation and reduction both involve simultaneously where one atom gets oxidised and the other gets reduced.
The equivalent weight of an oxidizing agent is that weight which accepts one mole electron in a chemical reaction.
$\text{Equivalent}\,\text{weight}\,\text{of}\,\text{an}\,\text{oxidant}\,\,\text{=}\,\text{ }\dfrac{\text{molecular}\,\text{weight}}{\text{number}\,\text{of}\,\text{electrons}\,\text{gained}\,\text{by}\,\text{one}\,\text{atom}}$
The equivalent weight of a reducing agent is the weight which donates one mole of electron in a chemical reaction.
$\text{Equivalent}\,\text{weight}\,\text{of}\,\text{reducant }\,\text{= }\dfrac{\text{molecular}\,\text{weight}}{\text{number}\,\text{of}\,\text{electrons}\,\text{lost}\,\text{by}\,\text{one}\,\text{atom}}$
In a redox reaction $\text{Equivalent}\,\text{weight}\,\,\text{= }\dfrac{\text{molecular}\,\text{weight}}{\text{change in oxidation}\,\text{number}}$
According to the law of equivalence, the substances combine together in the ratio of their equivalent masses.

& \text{Normality(N)}\,\,\text{=}\,\,\text{Molarity(M)}\,\,\text{ }\\!\\!\times\\!\\!\text{ }\,\,\text{valency}\,\text{factor(f)} \\\ & \text{For}\,\text{a redox reaction}\,\,-\text{valency}\,\text{factor(f)}=\,\,\text{change}\,\,\text{in}\,\text{oxidation}\,\text{number} \end{aligned}$$ **Complete Solution :** To calculate final oxidation state we will follow these steps – In first step we will calculate the gram equivalent of $\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$. The reaction in this question occurs in following manner- $$\text{KI}{{\text{O}}_{\text{3}}}\,\text{+}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}\,\,\to \,\text{SO}_{\text{4}}^{\text{2-}}\,\text{+}\,\,{{\text{I}}^{\text{x}}}$$ The oxidation state of $$\text{I}$$ in $\text{KI}{{\text{O}}_{\text{3}}}$ is $+5$ and the oxidation state of $\text{S}$ is changes from $+4\to +6$. So the valency factor (f) for $\text{S}\,\text{=}\,\text{2}$ So, $$\begin{aligned} & \text{No}\,\text{of}\,\text{gm}\text{.}\,\text{equivalent}\,\text{of}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}\text{ = }\,\text{N }\\!\\!\times\\!\\!\text{ }\,\,\text{Volume(L)} \\\ & =\,\text{M}\,\,\text{ }\\!\\!\times\\!\\!\text{ }\,\,\text{f}\,\text{ }\\!\\!\times\\!\\!\text{ }\,\,\text{Volume(L)} \\\ & \text{No}\,\text{of}\,\text{gm}\text{.}\,\text{equivalent}\,\text{of}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}=\,\dfrac{0.5\,\times \,2\,\times \,60}{1000}\,\,\,\,\,\\{\,\text{f = 2}\,\,\text{change}\,\,\text{in}\,\,\text{the oxidation}\,\text{state}\,\text{of}\,\text{sulphur }\\!\\!\\}\\!\\!\text{ } \\\ & \text{No}\,\text{of}\,\text{gm}\text{.}\,\text{equivalent}\,\text{of}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}=\,0.06\,\text{gm} \end{aligned}$$ So, $\text{KI}{{\text{O}}_{\text{3}}}$reacts with the 0.06gm equivalent of $\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$. In second step we will calculate the gram equivalent of $\text{KI}{{\text{O}}_{\text{3}}}$ $$\begin{aligned} & \text{No}\,\text{of}\,\text{mole}\,\text{of}\,\text{KI}{{\text{O}}_{\text{3}}}\,\,\,\,\,\,\text{n}\,\,=\,\,\dfrac{\text{W}}{\text{M}\text{.W}} \\\ & \text{n}\,\text{ = }\,\dfrac{2.14}{214}\,\,=\,\,0.01 \end{aligned}$$ $$\begin{aligned} & \text{Molarity}\,\text{of}\,\text{KI}{{\text{O}}_{\text{3}}} \\\ & \,\text{M}\,\text{=}\,\,\dfrac{\text{n}}{{{\text{V}}_{\text{(L)}}}} \\\ & \,\,\text{M}\,\text{=}\,\,\dfrac{\text{0}\text{.01}}{\text{100}\,\text{ml}}\text{ }\\!\\!\times\\!\\!\text{ }\,\text{1000} \\\ & \text{M}\,\,\text{=}\,\,\text{0}\text{.1} \end{aligned}$$ Since valency factor of $$\text{I}$$ = change in oxidation state = (5-x) Number of gm. Equivalent of $\text{KI}{{\text{O}}_{\text{3}}}$ $$\begin{aligned} & \text{gm}\,\text{eq}\text{.}\,\text{of}\,\text{KI}{{\text{O}}_{\text{3}}}\text{=}\,\text{N }\\!\\!\times\\!\\!\text{ }\,\,\text{Volume}\left( \text{L} \right) \\\ & \text{=}\,\text{M }\\!\\!\times\\!\\!\text{ }\,\text{f }\\!\\!\times\\!\\!\text{ }\,{{\text{V}}_{\text{(L)}}} \\\ & \,\text{=}\,\text{0}\text{.1 }\\!\\!\times\\!\\!\text{ }\,\,\text{(5-x) }\,\text{ }\\!\\!\times\\!\\!\text{ }\,\dfrac{\text{100}}{\text{1000}} \\\ & \text{gm}\,\text{eq}\text{.}\,\text{of}\,\text{KI}{{\text{O}}_{\text{3}}}\,\text{= }\,\text{0}\text{.01 }\\!\\!\times\\!\\!\text{ }\,\,\text{(5-x)} \\\ & \text{since, }\,\text{gram}\,\,\text{equivalent}\,\text{of}\,\text{reacting}\,\text{agent}\,\text{are}\,\text{equal} \\\ & \text{so,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{gm}\,\text{eq}\text{.}\,\text{of}\,\,\text{KI}{{\text{O}}_{\text{3}}}\text{=}\,\,\text{gm}\,\text{eq}\text{.}\,\text{of}\,\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}} \\\ & \text{0}\text{.01 }\\!\\!\times\\!\\!\text{ }\,\,\text{(5-x) = }\,\text{0}\text{.06} \\\ & \text{5-x}\,\,\text{=}\,\,\text{6} \\\ & \text{x = -1} \end{aligned}$$ So the value of final valency of $$\text{I}$$ in the redox reaction $$x = -1$$ **So, the correct answer is “Option B”.** **Note:** One equivalent of an element combined with one equivalent of another element, and in a chemical reaction equivalent and milli equivalent of reactants react in equal amounts to give same no. of equivalent or milli equivalent of products separately.