Answer
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Hint: In redox chemistry, a reducing agent is an element or compound which loses its electron to an electron-accepting species. It reduces the oxidizing agent and itself undergoes oxidation.
Complete answer:
Redox stands for the reduction and oxidation reactions.
-We know that reduction in the process of gaining and oxidation is the process of losing an electron to electron-accepting species. In a redox reaction, the charge of species decreases which accepts the electron and the charge on the species increases if it loses the electron.
-Reducing agents are the elements or compounds that lose an electron to the oxidizing agent. It undergoes the oxidation and reduces the other species.
-Characteristics of a Strong reducing agent are;
1) They easily donates their electron
2) They have low electronegativity
3) Relatively low ionization energies
4) They have a large atomic radius.
5) They have a high negative reduction potential
-The reduction potential or the redox potential is the measure of the behaviour of the species to gain or lose an electron to the electrode.
-If a solution has a higher redox potential it tends to accept the electron
-If the solution has low or negative redox potential it has a high tendency to lose an electron.
-We want to find out the best reducing agent among the given options.
-The reduction potential values for the$\text{A}{{\text{g}}^{\text{+}}}$,$\text{Z}{{\text{n}}^{\text{2+}}}$,$\text{C}{{\text{r}}^{\text{3+}}}$and ${{\text{I}}^{\text{-}}}$is given as follows,$\text{Zn}_{\text{(aq)}}^{\text{2+}}\text{+2}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{Z}{{\text{n}}_{\text{(s)}}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.7618V}$
$\text{Cr}_{\text{(aq)}}^{\text{3+}}\text{+}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{Cr}_{\text{(s)}}^{\text{2+}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.4248V}$
$\text{Ag}_{\text{(aq)}}^{\text{+}}\text{+}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{Ag}_{\text{(s)}}^{\text{2+}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.7996V}$
${{\text{I}}_{\text{2}}}_{\text{(s)}}\text{+2}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{2}{{\text{I}}^{\text{-}}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.5355V}$
-From the standard potential, we know that a larger negative value of the standard potential means that compounds or elements lose the electron easily.
-The standard potential $\text{A}{{\text{g}}^{\text{+}}}$has a higher negative value than the$\text{Z}{{\text{n}}^{\text{2+}}}\text{,C}{{\text{r}}^{\text{3+}}}\text{and }{{\text{I}}^{\text{-}}}$. Therefore $\text{A}{{\text{g}}^{\text{+}}}$is a stronger reducing agent.
So, the correct answer is “Option A”.
Additional Information:
In redox chemistry, the stronger the reducing agent, the weaker is its conjugate oxidizing agent. Fluorine gas is a strong oxidizing agent and ${{\text{F}}^{\text{-}}}$is a weak reducing agent. Similarly, weaker is the reducing agent stronger is the corresponding oxidizing agent. For example,
$\begin{align}
& {{\text{F}}_{\text{2}}}\text{(g) + 2}{{\text{e}}^{\text{-}}}\text{ }\to \text{ 2}{{\text{F}}^{\text{-}}} \\
& \text{(strong oxidising agent) (weakest reducing agent)} \\
\end{align}$
$\begin{align}
& \text{ N}{{\text{a}}^{\text{+}}}\text{ + }{{\text{e}}^{\text{-}}}\text{ }\to \text{ Na} \\
& \text{(Weakest oxidising agent) (Strongest reducing agent)} \\
\end{align}$
-Lithium is the strongest reduction agent because it has the smallest standard reduction potential. It decreases the charge on another species and itself oxidizes. It is one of the powerful reducing agents. It is smaller in size and loses electrons easily.
Note: Based on the electrode potential of metal, it can be anode in one reaction or cathode in other reactions. This can be positive or negative as seen on the voltmeter. For example, lead reduces the zinc, therefore in this pair lead acts as a cathode and zinc as an anode. However, lead is more oxidized than copper, therefore in $\text{Pb-Cu}$ pair copper acts as a cathode and lead as an anode.
Complete answer:
Redox stands for the reduction and oxidation reactions.
-We know that reduction in the process of gaining and oxidation is the process of losing an electron to electron-accepting species. In a redox reaction, the charge of species decreases which accepts the electron and the charge on the species increases if it loses the electron.
-Reducing agents are the elements or compounds that lose an electron to the oxidizing agent. It undergoes the oxidation and reduces the other species.
-Characteristics of a Strong reducing agent are;
1) They easily donates their electron
2) They have low electronegativity
3) Relatively low ionization energies
4) They have a large atomic radius.
5) They have a high negative reduction potential
-The reduction potential or the redox potential is the measure of the behaviour of the species to gain or lose an electron to the electrode.
-If a solution has a higher redox potential it tends to accept the electron
-If the solution has low or negative redox potential it has a high tendency to lose an electron.
-We want to find out the best reducing agent among the given options.
-The reduction potential values for the$\text{A}{{\text{g}}^{\text{+}}}$,$\text{Z}{{\text{n}}^{\text{2+}}}$,$\text{C}{{\text{r}}^{\text{3+}}}$and ${{\text{I}}^{\text{-}}}$is given as follows,$\text{Zn}_{\text{(aq)}}^{\text{2+}}\text{+2}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{Z}{{\text{n}}_{\text{(s)}}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.7618V}$
$\text{Cr}_{\text{(aq)}}^{\text{3+}}\text{+}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{Cr}_{\text{(s)}}^{\text{2+}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.4248V}$
$\text{Ag}_{\text{(aq)}}^{\text{+}}\text{+}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{Ag}_{\text{(s)}}^{\text{2+}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.7996V}$
${{\text{I}}_{\text{2}}}_{\text{(s)}}\text{+2}{{\text{e}}^{\text{-}}}\rightleftharpoons \text{2}{{\text{I}}^{\text{-}}}\text{ std}\text{.potential(}{{\text{E}}_{\text{0}}}\text{)=-0}\text{.5355V}$
-From the standard potential, we know that a larger negative value of the standard potential means that compounds or elements lose the electron easily.
-The standard potential $\text{A}{{\text{g}}^{\text{+}}}$has a higher negative value than the$\text{Z}{{\text{n}}^{\text{2+}}}\text{,C}{{\text{r}}^{\text{3+}}}\text{and }{{\text{I}}^{\text{-}}}$. Therefore $\text{A}{{\text{g}}^{\text{+}}}$is a stronger reducing agent.
So, the correct answer is “Option A”.
Additional Information:
In redox chemistry, the stronger the reducing agent, the weaker is its conjugate oxidizing agent. Fluorine gas is a strong oxidizing agent and ${{\text{F}}^{\text{-}}}$is a weak reducing agent. Similarly, weaker is the reducing agent stronger is the corresponding oxidizing agent. For example,
$\begin{align}
& {{\text{F}}_{\text{2}}}\text{(g) + 2}{{\text{e}}^{\text{-}}}\text{ }\to \text{ 2}{{\text{F}}^{\text{-}}} \\
& \text{(strong oxidising agent) (weakest reducing agent)} \\
\end{align}$
$\begin{align}
& \text{ N}{{\text{a}}^{\text{+}}}\text{ + }{{\text{e}}^{\text{-}}}\text{ }\to \text{ Na} \\
& \text{(Weakest oxidising agent) (Strongest reducing agent)} \\
\end{align}$
-Lithium is the strongest reduction agent because it has the smallest standard reduction potential. It decreases the charge on another species and itself oxidizes. It is one of the powerful reducing agents. It is smaller in size and loses electrons easily.
Note: Based on the electrode potential of metal, it can be anode in one reaction or cathode in other reactions. This can be positive or negative as seen on the voltmeter. For example, lead reduces the zinc, therefore in this pair lead acts as a cathode and zinc as an anode. However, lead is more oxidized than copper, therefore in $\text{Pb-Cu}$ pair copper acts as a cathode and lead as an anode.
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