Question

$K_2{Cr}_2{O}_7+HCl\to Cr{Cl}_3+H_{2}O+{Cl}_2+KCl$
Balance in basic medium by ion-electron identity spectator ion.

Answer 1

For the reaction

$K_{2}C{r}_2{O}_7+HCl\to CrC{l}_3+H_{2}O+C{l}_2+KCl$

Now for oxidation half

$2C{l}^{-}\to C{l}_2+2l^{-}$....(1)

for reduction half

$C{r}_2{O}_7^{2-}\to 2C{r}^{3+}$

Balancing

$C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

Balancing charge:

$C{r}_2{O}_7^{2-}+14H^{+}+6l^{-}\to 2C{r}^{3+}+7H_{2}O$...(2)

$3\times \left(1\right)+\left(2\right)$

$C{r}_2{O}_7^{2-}+14H^{+}+6C{l}^{-}\to 2C{r}^{3+}+3C{l}_2+6l^{-}+7H_{2}O$

Writing in molecular form.

$K_{2}C{r}_2{O}_7+6HCl+8HCl\to 2CrC{l}_3+3C{l}_2+2KCl+7H_{2}O$

$K_{2}C{r}_2{O}_7+14HCl\to 2CrC{l}_3+3C{l}_2+2KCl+7H_{2}O$

for basic medium

$K_{2}Cr{O}_7+14HCl+14NaOH\to 2CrC{l}_3+3C{l}_2+2KCl+7H_{2}O+14NaOH$

$K_{2}C{r}_2{O}_7+14NaCl+7H_{2}O\to 2CrC{l}_3+3C{l}_2+2KCl+14NaOH$