Question

Balance the following equations by ion electron method:
${C{r}_2{O}_7}^{2+}+S{O}_2+H^{+}\to C{r}^{3+}+{HS{O}_4}^{-}+H_{2}O$.

Answer 1

$C{r}_2{O}_7^{2-}+S{O}_2+H^{+}\to C{r}^{3+}+HS{O}_4^{-}+H_{2}O$
$1$st step: Splitting into two half reactions,
$\underset{\left(Reductionhalfreaction\right)}{C{r}_2{O}_7^{2-}+H^{+}\to C{r}^{3+}+H_{2}O}$; $\underset{\left(Oxidationhalfreaction\right)}{S{O}_2\to HS{O}_4^{-}}$
$2$nd step: Adding $H^{+}$ ions to side deficient in hydrogen,
$C{r}_2{O}_7^{2-}+14H^{+}\to C{r}^{3+}+7H_{2}O$
$3$rd step: Adding water to the side deficient in oxygen,
$S{O}_2+2H_{2}O\to HS{O}_4^{-}+3H^{+}$
$4$th step: Making atoms equal on blth sides,
$C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$
$5$th step: Adding electrons to the sides deficient in electrons,
$C{r}_2{O}_7^{2-}+14H^{+}+6e\to 2C{r}^{3+}+7H_{2}O$;
$S{O}_2+2H_{2}O\to HS{O}_4^{-}+3H^{+}+2e$
$6$th step: Balancing electrons in both the half reactions,
$C{r}_2{O}_7^{2-}+14H^{+}+6e\to 2C{r}^{3+}+7H_{2}O$
$[S{O}_2+2H_{2}O\to HS{O}_4^{-}+3H^{+}+2e]\times 3$
$7$th step: Adding both the half reactions,
$C{r}_2{O}_7^{2-}+5H^{+}+3S{O}_2\to 2C{r}^{3+}+3HS{O}_4^{-}+H_{2}O$.