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Hint: Standard oxidation potential is the potential of the electrode when it undergoes the oxidation. Reducing agent is the chemical species which itself undergoes the oxidation and reduces the other species. Thus the species which have higher standard oxidation potential (SOP) can readily lose the electron to reduce the other chemical species and therefore it is a strong reducing agent.
Complete step by step answer:
The standard oxidation potential is the tendency of a species to undergo the oxidization or the tendency of an electrode to lose an electron. It is measured in volts. The standard oxidation potential for a half-reaction can be written as:
$\text{ A(s)}\to {{\text{A}}^{\text{m+}}}\text{+m}{{\text{e}}^{\text{-}}}$
For example, the standard oxidation potential that is easy to lose an electron for copper as$\text{Cu(s)}\to \text{C}{{\text{u}}^{\text{2+}}}\text{+2}{{\text{e}}^{\text{-}}}$
The standard oxidation potential is $\text{E}_{\text{O}}^{\text{0}}\text{(SOP)= - 0}\text{.345 V}$
That is higher the positive value of oxidation potential higher is the tendency to lose an electron.
We know that the reducing agent is the chemical species which reduces the other species present in the reaction and itself undergoes the oxidation. Thus the species which have a higher standard oxidation potential is a strong reducing agent.
In the above problem, we are provided with the following standard oxidation potentials. $-0.55\text{ , +0}\text{.86 , -0}\text{.21 , +0}\text{.38}$
Let's arrange all the values in ascending order
$-0.55\text{ }<\text{ -0}\text{.21 }<\text{ +0}\text{.38 }<\text{+0}\text{.86}$
Here, $\text{+0}\text{.86 V}$is the highest standard oxidation potential .thus the chemical species which have the $\text{+0}\text{.86 V}$as the SOP is the strong reducing agent among the given values.
Hence, (B) is the correct option.
Note: one should not confuse between the standard oxidation potential (SOP) and standard reduction potential (SRP).The reduction potential is the measure to accept the electron. Thus the standard oxidation potential and the standard reduction potential are related oppositely to each other for the same chemical species. That is $\text{E}_{\text{O}}^{\text{0}}\text{(SRP)= - E}_{\text{O}}^{\text{0}}\text{(SOP)}$
Complete step by step answer:
The standard oxidation potential is the tendency of a species to undergo the oxidization or the tendency of an electrode to lose an electron. It is measured in volts. The standard oxidation potential for a half-reaction can be written as:
$\text{ A(s)}\to {{\text{A}}^{\text{m+}}}\text{+m}{{\text{e}}^{\text{-}}}$
For example, the standard oxidation potential that is easy to lose an electron for copper as$\text{Cu(s)}\to \text{C}{{\text{u}}^{\text{2+}}}\text{+2}{{\text{e}}^{\text{-}}}$
The standard oxidation potential is $\text{E}_{\text{O}}^{\text{0}}\text{(SOP)= - 0}\text{.345 V}$
That is higher the positive value of oxidation potential higher is the tendency to lose an electron.
We know that the reducing agent is the chemical species which reduces the other species present in the reaction and itself undergoes the oxidation. Thus the species which have a higher standard oxidation potential is a strong reducing agent.
In the above problem, we are provided with the following standard oxidation potentials. $-0.55\text{ , +0}\text{.86 , -0}\text{.21 , +0}\text{.38}$
Let's arrange all the values in ascending order
$-0.55\text{ }<\text{ -0}\text{.21 }<\text{ +0}\text{.38 }<\text{+0}\text{.86}$
Here, $\text{+0}\text{.86 V}$is the highest standard oxidation potential .thus the chemical species which have the $\text{+0}\text{.86 V}$as the SOP is the strong reducing agent among the given values.
Hence, (B) is the correct option.
Note: one should not confuse between the standard oxidation potential (SOP) and standard reduction potential (SRP).The reduction potential is the measure to accept the electron. Thus the standard oxidation potential and the standard reduction potential are related oppositely to each other for the same chemical species. That is $\text{E}_{\text{O}}^{\text{0}}\text{(SRP)= - E}_{\text{O}}^{\text{0}}\text{(SOP)}$
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