Question

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

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (C)

6

For a balanced chemical reaction, the total charge on the reactant side must be equal to the total charge on the product side.

$\therefore$ Total charge on reactant side $=$ total charge on the product side

$\Rightarrow (-2)+14\times (+1)+n\times (+2)=2\times (3+)+n\times (3+)$

$\Rightarrow$ $-2+14+2n=6+3n$

$\Rightarrow$ $n=6$

Hence, the value of $n$ for the given reaction is 6.