Question

The value of ‘n’ in the reaction $Cr _{2} O _{7}^{2-}+14 H ^{+}+n Fe ^{2+} \longrightarrow 2 Cr ^{3+} +n Fe ^{3+} +7 H _{2} O$ will be

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (C)

$Cr _{2} O _{7}^{2-}+14 H ^{+}+n Fe ^{2+} \longrightarrow 2 Cr ^{3+}+n Fe ^{3+}+7 H _{2} O$
$Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e^{-} \longrightarrow 2 Cr ^{3+}+7 H _{2} O
\text { (reduction) }$ ...(i)
$Fe ^{2+} \longrightarrow Fe ^{3+}+e^{-} \text {(oxidation) }$ ...(ii)
Eq. (ii) is multiplied by 6
$6 Fe ^{2+} \longrightarrow 6 Fe ^{3+}+6 e^{-}$
Thus, balanced equation is
$Cr _{2} O _{7}^{2-}+14 H ^{+}+6 Fe ^{2+} \longrightarrow $
$2 Cr ^{3+}+6 Fe ^{3+}+7 H _{2} O$
Hence, the value of ' $n$ ' is $6$.