Question

Represent the following equation in ionic form.
$K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6FeS{O}_4\to 3F{e}_2(S{O}_4{)}_3+C{r}_2(S{O}_4{)}_3+7H_{2}O+K_{2}S{O}_4$.

Answer 1

In this equation except $H_{2}O$, all are ionic in nature. Representing these compounds in ionic forms,
$2K^{+}+C{r}_2{O}_7^{2-}+14H^{+}+7S{O}_4^{2-}+6F{e}^{2+}+6S{O}_4^{2-}\to 6F{e}^{3+}+9S{O}_4^{2-}+2C{r}^{3+}+3S{O}_4^{2-}+2K^{+}+S{O}_4^{2-}+7H_{2}O$
$2K^{+}$ ions and $13S{O}_4^{2-}$ ions are common on both sides, so these are cancelled. The desired ionic equation reduces to the following equation
$C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}=6F{e}^{3+}+2C{r}^{3+}+7H_{2}O$
Total charges are equal on both sides; thus, the balanced ionic equation is the same as above.