Question

Write balanced half reactions for the following redox reaction:
$C{r}_2{O}_7^{2-}+F{e}^{2+}\to C{r}^{3+}+F{e}^{3+}$

• A. Reduction: $6e^{-}+14H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+7H_{2}O$
  Oxidation: $F{e}^{2+}\to F{e}^{3+}+e^{-}$
• B. Oxidation: $6e^{-}+14H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+7H_{2}O$
  Reduction: $F{e}^{2+}\to F{e}^{3+}+e^{-}$
• C. Reduction: $4e^{-}+14H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+7H_{2}O$
  Oxidation: $F{e}^{2+}\to F{e}^{3+}+e^{-}$
• D. Oxidation: $6e^{-}+14H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+6H_{2}O$
  Reduction: $F{e}^{2+}\to F{e}^{3+}+e^{-}$

Answer 1

The correct option is A Reduction: $6e^{-}+14H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+7H_{2}O$

Oxidation: $F{e}^{2+}\to F{e}^{3+}+e^{-}$
Given Redox reaction is

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

Oxidation half reaction:- $F{e}^{+2}⟶F{e}^{+3}+e^{-}$

[as oxidation number increases from $2$ to $3$]

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

[as oxidation number decreases from $6$ to $3$]