Question

Choose the reactions which are not feasible
1.Iron + zinc sulphate $ \to $Iron sulphate + Zinc
2.Magnesium + silver nitrate $\to$ magnesium nitrate + silver
3.copper + dil.sulphuric acid $\to$ copper sulphate + $H_2$
4.Zinc + Ferrous sulphate $\to$ Iron + Zinc sulphate
A. I,II and III
B. III and IV
C. I and III only
D. All of these.

Answer 1

Hint: The values present below represent the standard electrode potential. The lower the SRP values the better the reactant is at undergoing reduction. Therefore, in the reactions the oxidizing agents must have lower or more negative standard electrode potential. The opposite is true for reducing agents.
\[Li/L{i^ + } \to - 3.03\]
\[Z{n^{ + 2}}/Zn \to - 0.76\]
\[F{e^{ + 2}}/Fe \to - 0.44\]
\[{H^ + }/{H_2} \to 0\]
\[C{u^{ + 2}}/Cu \to + 0.34\]

Complete step by step answer:
As mentioned in the hint, using the standard electrode potential values of the series we can classify the reactants into oxidizing agents and reducing agents.
It is important to remember that negative standard electrode potential as in the case of Lithium denotes that it is a good reducing agent. This means that it will undergo oxidation very easily and hence we can consider the electrode containing Li to be the anode.
The lower and more positive values represent the ability of the reactant to undergo reduction. That means that Au in this series can act as a good oxidizing agent.
In this equation, \[Fe\] is in ground state or the oxidation number of Iron can be considered to be $0$ . Whereas \[Zn\] is in $ + 2$ oxidation state. This means that upon reacting, $Fe$ changes from $0$ to $ + 2$ . For this reaction to be feasible, value of Standard reduction potential of \[Fe\] has to be more positive than \[Zn\] and the value is more positive as SRP of \[Fe\] is $ - 0.44$ and value of \[Zn\] is $ - 0.76$ . Thus, the first reaction is not feasible.
In this second reaction, $Mg$ is anode and undergoes oxidation. Thus, it should have higher Standard reduction potential value than $Ag$ . But it does not therefore, this reaction is not feasible.
$Cu$ is the anode here and undergoes oxidation from $0 \to + 2$ and Standard reduction potential value is more than hydrogen. This means that this reaction is possible.
\[Zn\] undergoes oxidation and changes its oxidation state from $0 \to + 2$ . \[Zn\] can act as the anode with ease as it has Standard reduction potential that is more negative or lesser than \[Fe\] . Therefore, this reaction is feasible.

So, the correct answer is Option B.

Note: The lower the Standard reduction potential value the higher the tendency of the electrode to undergo oxidation. That means this electrode will be a better reducing agent. This electrode will be considered to be the anode.