Question

(a) How will you show that iron is more reactive than copper?
(b) Explain how a metal which is high up in the reactivity series obtained from its compound. Give equations also.

Answer 1

Hint: A more reaction metal can displace a less reactive metal from a solution. Reactivity series arranges the elements in decreasing order of their standard reduction potential (i.e. in order of increasing reactivity). Standard reduction potential of iron ($F{{e}^{2+}},Fe$) is -0.44 V while the standard reduction potential of copper ($C{{u}^{2+}},Cu$) is +0.34 V.

Complete step by step answer:
(a) Standard oxidation and standard reduction potentials are equal in magnitude but opposite in sign. Therefore, the oxidation potential of iron ($Fe,F{{e}^{2+}}$) = 0.44 V and the oxidation potential of copper ($Cu,C{{u}^{2+}}$) is +0.34 V.
It is clear from the above values that the standard oxidation potential of iron is more than that of copper. It means that iron can more easily lose electrons and get oxidized. Hence, it is more reactive than copper.
We can prove this with the help of the following activity.
Take a solution of copper sulphate ($CuS{{O}_{4}}$) in a beaker. It is blue in colour.
Dip a clean iron nail in the solution of $CuS{{O}_{4}}$
Now wait for a few minutes to observe the change.

We will see that the blue $CuS{{O}_{4}}$ solution turns light green and a brown coating will form on the surface of the iron nail,
The changes observed are due to the fact that Cu has been displaced by Fe. The change in colour is due to the formation of ferrous sulphate $FeS{{O}_{4}}$ which is light green in colour. Hence, we can say that iron being more reactive than copper displaces less reactive copper in a solution.

(b) Standard reduction potential of copper ($C{{u}^{2+}},Cu$), i.e. +0.34 V. is more than that of iron ($F{{e}^{2+}},Fe$), i.e. -0.44 V. Thus, Cu is present higher up in the activity series than Fe. Therefore, Cu can undergo reduction more easily than Fe.
In other words, we can say that Cu will act as an oxidizing agent and Fe will act as a reducing agent. Fe will reduce Cu to $C{{u}^{2+}}$ and Fe itself will get oxidized to $F{{e}^{2+}}$. We already saw that in the above activity that Fe can displace Cu from its salt solution.

The reactions that occur during the whole process are given below:
Oxidation half reaction : $Fe\to F{{e}^{2+}}+2{{e}^{-}}$
Reduction half reaction: $C{{u}^{2+}}+2{{e}^{-}}\to Cu$

The overall reaction showing that Cu which is high up in the reactivity series is obtained from its compound $CuS{{O}_{4}}$ is given below:
     \[Fe+CuS{{O}_{4}}\to FeS{{O}_{4}}+Cu\]

Note: Keep in mind that Reduction potential is the tendency of an electrode to undergo reduction. Oxidation potential is the tendency of an electrode to lose electrons and get oxidized. Higher the oxidation potential, greater is the reactivity.