Question

Balance this equation.

$C{r}_2{O}_7^{2-}+S{O}_3^{2-}\to S{O}_4^{2-}+C{r}^{3+}$

Answer 1

Given,

$C{r}_2{O}_7^{2-}+S{O}_3^{2-}⟶S{O}_4^{2-}+C{r}^{3+}$

In acidic medium

$C{r}_2^{+6}{O}_7^{2-}⟶2C{r}^{3+}$ [Reduction half reaction]

$C{r}_2{O}_7^{2-}⟶2C{r}^{3+}+7H_{2}O$ [balance oxygens]

$C{r}_2{O}_7^{2-}+14H^{+}⟶2C{r}^{3+}+7H_{2}O$ [balance hydrogen]

$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2C{r}^{3+}+7H_{2}O$ [balance charge]

$S^{+4}{O}_3^{2-}⟶S^{+6}{O}_4^{2-}$ [Oxidation half reaction]

$S{O}_3^{2-}+2H_{2}O⟶S{O}_4^{2-}$ [balance oxygen]

$S{O}_3^{2-}+2H_{2}O⟶S{O}_4^{2-}+4H^{+}$ [balance hydrogen]

$S{O}_3^{2-}+2H_{2}O⟶S{O}_4^{2-}+4H^{+}+4e^{-}$ [balance charge]

Add both by balancing charge of electrons.

$3S{O}_3^{2-}+6H_{2}O⟶3S{O}_4^{2-}+12H^{+}+12e^{-}1$

$2C{r}_2{O}_7^{2-}+28H^{+}+12e^{-}⟶4C{r}^{3+}+14H_{2}O2$

$⟹2C{r}_2{O}_7^{2-}+3S{O}_3^{2-}+28H^{+}+6H_{2}O⟶4C{r}^{3+}+3S{O}_4^{2-}+14H_{2}O+12H^{+}$

$⟹2C{r}_2{O}_7^2+3S{O}_3^{2-}+16H^{+}⟶4C{r}^{3+}+8H_{2}O+3S{O}_4^{2-}$