Question

In a reaction, $Cr _{2} O _{7}^{2-}$ is reduced to $Cr ^{3+}$. What will be the concentration of $0.1\, M\, K _{2} Cr _{2} O _{7}$ in equivalent per litre? $Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O$

• A. $0.9\,N$
• B. $0.6 \, N$
• C. $0.3 \, N$
• D. $0.2 \,N$

Answer 1

Answer: (B)

$N = M \times$ Valence factor
$\because \mathrm{N}=0.1 \times 6\left(2 \mathrm{Cr}^{6+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}\right)$
$K _{2} Cr _{2} O _{7}=0.6\, N$