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:
Correct Option: , 2
Solution:
Higher the reduction potential, higher will be oxidising power. So,
$\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$