Question

The value of ‘n’ in the reaction $C{r}_2{O}_7^{2-}+14H^{+}+nF{e}^{2+}⟶2C{r}^{3+}+nF{e}^{3+}+7H_{2}O$ will be

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (C)

$C{r}_2{O}_7^{2-}+14H^{+}+nF{e}^{2+}⟶2C{r}^{3+}+nF{e}^{3+}+7H_{2}O$
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2C{r}^{3+}+7H_{2}O<br/>\text{ (reduction) }$ ...(i)
$F{e}^{2+}⟶F{e}^{3+}+e^{-}\text{(oxidation) }$ ...(ii)
Eq. (ii) is multiplied by 6
$6F{e}^{2+}⟶6F{e}^{3+}+6e^{-}$
Thus, balanced equation is
$C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}⟶$
$2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$
Hence, the value of ' $n$ ' is $6$.