Question

Balance the following redox reaction.
$(C{r}_2{O}_7{)}_{\left(aq\right)}^{2-}+F{e}_{\left(aq\right)}^{2+}\to C{r}_{\left(aq\right)}^{3+}+F{e}_{\left(aq\right)}^{3+}$ (acidic medium)

Answer 1

$F{e}^{2+}+C{r}_2{O}_7^{2-}+H^{+}⟶F{e}^{3+}+C{r}^{3+}+H_{2}O$

(1) Oxidation: $F{e}^{2+}\to F{e}^{3+}$

Reduction: ${\stackrel{+6}{Cr}}_2{O}_7^{2-}⟶\stackrel{+3}{C}{r}^{3+}$

(2) Balancing the atoms

$F{e}^{2+}⟶F{e}^{3+}$

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

(3) Balance the charge

$F{e}^{2+}⟶F{e}^{3+}+e^{-}⟶\left(1\right)$

$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2C{r}^{3+}+7H_{2}O⟶\left(2\right)$

(4) Equation (1) $\times 6+$ Equation (2) $\times 1$ Balancing electrons

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

(5) Simplifying;

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