On balancing the given redox reaction, the coefficients $a$, $b$ and $c$ are found to be, respectively: \ce{a$Cr2O{7}^2-(aq)+$b$SO{3}^2-(aq)+$c$H^{+}(aq)->2C{r}^3+(aq)+$b$SO{4}^2-(aq)+$c/2 H2O(l)}

Solution & Explanation

\textbf{Answer:} (a) $1,3,8$ \textbf{Solution:} Balanced reaction: $\mathrm{Cr}{}_2\mathrm{O}{}_7{}^{2-}+3\mathrm{SO}{}_3{}^{2-}+8\mathrm{H}{}^{+}\to 2\mathrm{Cr}{}^{3+}+3\mathrm{SO}{}_4{}^{2-}+4\mathrm{H}{}_2\mathrm{O}$. Cr is reduced from $+6$ to $+3$ (2 Cr gain $6e^{-}$); S is oxidised from $+4$ to $+6$ (each loses $2e^{-}$, so 3 sulphite ions lose $6e^{-}$), balancing electron transfer. Matching with the given form, $a=1$, $b=3$, $c=8$. Hence (a).