7. Redox reactions — Chemistry STD 11 Science — Question

$Cr _2 O _7^{2-}+ Sn ^{+2}+ H ^{+} \rightarrow Sn ^{-4}+ Cr ^{+3}+ H _2 O$

Step $1:$ Assign the oxidation state using oxidation number of $O =-2$.

we get:

$Cr _2 O _7^{2-}+ Sn ^{+2+}+ H ^{+} \rightarrow Sn ^{+4}+ Cr ^{+3}+ H _2 O$

Reduction half-reaction:

$Cr _2 O _7^{2-} \rightarrow 2 Cr ^{-3}:$ Gain of 6 electrons

Oxidation half reaction:

$Sn ^{+2} \rightarrow Sn ^{+4}:$ Loss of $2$ electrons

Step $2:$ Equalise the number of electrons as:

Oxidation half reaction:

$3 Sn ^{-2} \rightarrow 3 Sn ^{+4}$

Step $3:$ balance $O$ atoms by adding $H _2 O$ and then $H$ by $H ^{+}$

$Cr _2^{+6} O _7^{2-}+14 H ^{+} \rightarrow 2 Cr ^{-3+}+7 H _2 O$

Step $4:$ overall reaction:

$Cr _2^{+6} O _7^{2-}+14 H ^{+}+3 Sn ^{+2} \rightarrow 2 Cr ^{3+}+3 Sn ^{+4}+7 H _2 O$

Thus $3$ mole of $Sn ^{2+}$ will reduce 1 moles of $K _2 Cr _2 O _7$.

Therefore,$1$ mole of $Sn ^{2+}$ will reduce $\frac{1}{3}$ moles of $K _2 Cr _2 O _7$.