Chemistry - Redox Reactions Question with Solution | TestHub

Electrons supplied per mole of reductant: $\mathrm{FeS}{\mathrm{O}}_4$ = 1; $\mathrm{N}{\mathrm{a}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ or ${\mathrm{K}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ $\to$ $2C{\mathrm{O}}_2$ = 2; ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_3$ $\to$ ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4$ = 2. (A) $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}$ has 2 Cr(+6) $\to$ 2 $\mathrm{C}{\mathrm{r}}^{3+}$, needing 6 ${\mathrm{e}}^{-}$: 3.0 mol ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_3$ supplies 6 (S). (B) Mn(+7) $\to$ $\mathrm{M}{\mathrm{n}}^{2+}$ needs 5 ${\mathrm{e}}^{-}$: 2.5 mol $\mathrm{N}{\mathrm{a}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ supplies 5 (Q). (C) $\mathrm{Mn}{\mathrm{O}}_2$ $\to$ $\mathrm{M}{\mathrm{n}}^{2+}$ needs 2 ${\mathrm{e}}^{-}$: 1.0 mol ${\mathrm{K}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ (R). (D) [$\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$ $\to$ [$\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$ needs 1 ${\mathrm{e}}^{-}$: 1.0 mol $\mathrm{FeS}{\mathrm{O}}_4$ (P).