Embibe Experts Solutions for Chapter: Redox Reactions, Exercise 1: Exercise 1

The oxidation number of sulphur is $+4$ in

A redox reaction among the following is:

(i) ${\mathrm{CdCl}}_2+2\mathrm{KOH}\to \mathrm{Cd}(\mathrm{OH}{)}_2+2\mathrm{KCl}$

(ii) ${\mathrm{BaCl}}_{2 }+{\mathrm{K}}_2{\mathrm{SO}}_4\to {\mathrm{BaSO}}_4+2\mathrm{KCI}$

(iii) ${\mathrm{CaCO}}_3\to \mathrm{CaO}+{\mathrm{CO}}_2$

(iv) $2\mathrm{Ca}+{\mathrm{O}}_2\to 2\mathrm{CaO}$

The metal with the highest oxidation state present in ${\mathrm{K}}_2{\mathrm{CrO}}_4, {\mathrm{NbCl}}_5$ and ${\mathrm{MnO}}_2$ is:

The number of electrons required to reduce chromium completely in ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$ to ${\mathrm{Cr}}^{3+}$ in acidic medium, is

In balancing the reaction, ${\mathrm{xH}}_2\mathrm{S}+2{\mathrm{NaNO}}_3+2\mathrm{HCl}\to \mathrm{y} \mathrm{S}+\mathrm{zNO}+\mathrm{kNaCl}+4{\mathrm{H}}_2\mathrm{O}$, one would get $\mathrm{x}, \mathrm{y}, \mathrm{z}$ and $\mathrm{K},$ respectively, as

In the reaction ${\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{S}⟶3\mathrm{S}+2{\mathrm{H}}_2\mathrm{O},$ the substance that is oxidised is:

The reaction, ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+\mathrm{m} {\mathrm{FeSO}}_4+\mathrm{n} {\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+\mathrm{p} {\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{K}}_2{\mathrm{SO}}_4+\mathrm{q} {\mathrm{H}}_2\mathrm{O}$ when balanced, $\mathrm{m}, \mathrm{n}, \mathrm{p},$ and $\mathrm{q}$ are, respectively:

The oxidation number of sulphur in perdisulphuric acid is: