Question

The more positive the value of ${{E}^{\circ }}$, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below, find out which of the following is the strongest oxidizing agent. ${{E}^{\circ }}$ values:
$\begin{align}
  & F{{e}^{3+}}/F{{e}^{2+}}=+0.77V\text{ , }{{\text{I}}_{2(s)}}/{{I}^{-}}=+0.54V \\
 & C{{u}^{2+}}/Cu=+0.34V\text{ , A}{{\text{g}}^{+}}/Ag=+0.80V \\
\end{align}$
(A) $F{{e}^{3+}}$
(B) ${{I}_{2(s)}}$
(C) $C{{u}^{2+}}$
(D) $A{{g}^{+}}$

Answer 1

Hint: Oxidising agent is a redox reagent which itself gets reduced and oxidized other compounds in the redox reaction. Standard potential of the cell is the reduction potential of the cell.

Complete step by step answer:
- Here, we are being asked to find the strongest oxidizing agent from the given options.
- We know that oxidizing agent is the agent which oxidizes other compounds and itself gets reduced in the redox reaction. So, the agent which will have the highest reduction potential will have the highest oxidizing strength.
- We are also given that more the value of ${{E}^{\circ }}$, higher the tendency of it to get reduced. Here, ${{E}^{\circ }}$ is the standard reduction potential. Standard reduction potential of the cell is the reduction potential of the cell.
- So, in order to compare their oxidizing strength, we will compare the standard potential of the half cells and we can say that higher the value of the standard potential of the half cell, higher the oxidizing power will be.

Now, we are given that
$\begin{align}
  & F{{e}^{3+}}/F{{e}^{2+}}=+0.77V\text{ , }{{\text{I}}_{2(s)}}/{{I}^{-}}=+0.54V \\
 & C{{u}^{2+}}/Cu=+0.34V\text{ , A}{{\text{g}}^{+}}/Ag=+0.80V \\
\end{align}$

Here, we can see that the potentials of all half cells are positive values. That means all of them can act as oxidizing agents. Out of them the standard potential of $A{{g}^{+}}/Ag$ is the highest which is +0.80V. So, out of given four half cells, $A{{g}^{+}}/Ag$ will have highest oxidizing power.
So, the correct answer is “Option D”.

Note: Remember that we need not to multiply the number of electrons involved in the half cell reaction with the potential of the cell in order to measure its oxidizing strength. Do not forget that oxidizing agents oxidize other compounds and reducing agents reduce other compounds.