Question

Given:
\[C{o^{3 + }} + {e^ - } \to C{o^{2 + }};{E^o} = + 1.81V\]
\[P{b^{4 + }} + 2{e^ - } \to P{b^{2 + }};{E^o} = + 1.67V\]
\[C{e^{4 + }} + {e^ - } \to C{e^{3 + }};{E^o} = + 1.61V\]
\[B{i^{3 + }} + 3{e^ - } \to Bi;{E^o} = + 0.20V\]

The oxidizing power of the species will increase in the order:
(A) \[C{e^{4 + }} < P{b^{4 + }} < B{i^{3 + }} < C{o^{3 + }}\]
(B) \[C{o^{3 + }} < P{b^{4 + }} < C{e^{4 + }} < B{i^{3 + }}\]
(C) \[C{o^{3 + }} < C{e^{4 + }} < B{i^{3 + }} < P{b^{4 + }}\]
(D) \[B{i^{3 + }} < C{e^{4 + }} < P{b^{4 + }} < C{o^{3 + }}\]

Answer 1

Hint: The larger the value of the standard reduction potential for an element, the easier will be for the element to be reduced, hence better will the oxidizing agent.

Complete answer:
-The measure of the tendency of a molecule to be reduced by taking up new electrons is given by reduction potential of the element.
-The reduction potential of a molecule under standard conditions is known as the standard reduction potential.
-Standard reduction potentials are useful in determining the directionality of a reaction.
-The strengths of oxidizing and reducing agents are given by their standard oxidation potential and standard reduction potential respectively.
-An electrochemical cell has two types of half-reactions, an oxidation half-reaction and a reduction half-reaction. A reduced species is formed when oxidizing species accepts electrons as a result of which oxidation occurs at the anode. When an oxidized species loses electrons, a reduced species is formed as a result of reduction at the cathode.
\[{\rm{Reduced~ species }} \to {\rm{ Oxidized species + n}}{{\rm{e}}^ - }{\rm{ (oxidation at anode)}}\]
\[{\rm{Oxidized~ species + n}}{{\rm{e}}^ - }{\rm{ }} \to {\rm{Reduced species (reduction at cathode)}}\]

-The cell potential of a cell is the sum of oxidation potential and the reduction potential of the cell.
-Each half-cell is associated with an electrode-solution potential, whose magnitude depends on the nature of a particular electrode reaction and the concentration of the dissolved electroactive species. The sign of this potential difference depends on the direction in which the electrode reaction proceeds (either oxidation or reduction).
-The more positive the half-cell EMF, the greater will be the tendency of the reductant to donate electrons, and the smaller will be the tendency of the oxidant to accept electrons.
-Reduction potential of a substance is the ability of the substance to be reduced. So, with the increase in the reduction potential, reducing ability also increases.
-The ability of the substance to make other substances lose electrons is known as the oxidizing power. So with the increase in reduction potential, the oxidizing power of the substance also increases.
-Therefore, the correct increasing order of oxidizing power is given as-
\[B{i^{3 + }} < C{e^{4 + }} < P{b^{4 + }} < C{o^{3 + }}\]

Hence, the correct answer is option \[B{i^{3 + }} < C{e^{4 + }} < P{b^{4 + }} < C{o^{3 + }}\]
So, the correct answer is “Option D”.

Note: Standard electrode potential cannot be determined empirically. For measuring the electrode potential of a cell, the unknown electrode is coupled with a reference electrode SHE (Standard Hydrogen Electrode) whose s potential is defined and is set to 0.00V. SHE along with unknown electrode is paired to form a galvanic cell and the galvanic cell gives the potential of the unknown electrode.