# oxidizing power of the species will increase in the order:

Question:

Given:

$\mathrm{Co}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Co}^{2+} ; \mathrm{E}^{\mathrm{o}}=+1.81 \mathrm{~V}$

$\mathrm{~Pb}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{2+} ; \mathrm{E}^{\mathrm{o}}=+1.67 \mathrm{~V}$

$\mathrm{Ce}^{4+}+\mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+} ; \mathrm{E}^{0}=+1.61 \mathrm{~V}$

$\mathrm{Bi}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Bi} ; \mathrm{E}^{\mathrm{o}}=+0.20 \mathrm{~V}$

oxidizing power of the species will increase in the order:

1. $\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Co}^{3+}$

2. $\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$

3. $\mathrm{Co}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Pb}^{4+}$

4. $\mathrm{Co}^{3+}<\mathrm{Pb}^{4+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}$

Correct Option: , 2

Solution:

Higher the reduction potential, higher will be oxidising power. So,

$\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$