Given :

Question:

Given :

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

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

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

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

Oxidizing power of the species will increase in the order :

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

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

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

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

Correct Option: , 4

Solution:

$\mathrm{E}_{\text {Red }}^{\circ} \uparrow \Rightarrow$ oxidizing power $\uparrow$

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