Chemistry - Redox Reactions Question with Solution | TestHub

Key Idea (i) Sum of oxidation states of all atoms charge of ion. (ii) Oxidation number of oxygen $=-2$. Let the oxidation state of $\mathrm{P}$ in ${\mathrm{PO}}_4^{3-}$ is $x$. $$\therefore \begin{array}{rl}& {\mathrm{PO}}_4^{3-}\\ & \therefore \mathrm{x}+4(-2)=-3\\ & \mathrm{x}-8=-3\\ & \mathrm{x}=+5\end{array}$$ Let the oxidation state of $\mathrm{S}$ in ${\mathrm{SO}}_4^{2-}$ is . $$\begin{array}{rl}& {\mathrm{SO}}_4^{2-}\\ & \therefore \mathrm{y}+4(-2)=-2\\ & \mathrm{y}-8=-2\\ & \mathrm{y}=+6\end{array}$$ Let the oxidation state of $\mathrm{Cr}$ in ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ is . $$\therefore \begin{array}{rl}& {\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\\ & 2\times \mathrm{z}+7(-2)=-2\\ & 2\mathrm{z}-14=-2\\ & \mathrm{z}=+6\end{array}$$ Hence, oxidation state of $\mathrm{P},\mathrm{S}$ and $\mathrm{Cr}$ are +5 , +6 and +6 respectively.