Question

Chemical equation for the reaction between CuSO$_4$ solution and an iron nail is given below.
CuSO$_4$ + Fe $\rightarrow$ FeSO$_4$ + Cu
a) Write the reduction reaction taking place here.
b) Give reason for the displacement of Cu by Fe from CuSO$_4$ solution.

Answer 1

Hint: In this chemical equation observe the loss, and gain of electrons. Identify the substance acting as reducing agent, and oxidizing agent; and displacement is based on the electrochemical series. Now, the reduction reaction in part (a), and reason in part (b) could be known.

Complete step by step answer:
*First, we will discuss the part (a), we have to write the reduction reaction.
*As we know that in terms of electrons; reduction is based on the gain of electrons, and the electron acceptor is termed as an oxidizing agent.
*Thus, in the chemical equation, we can see that in the copper sulphate, copper exists in +2 oxidation state.
*Then, to form Cu it will reduce, and there will be gain of two electrons.
*Therefore, the reduction reaction can be written as:
Cu$^{2+}$ + 2e$^{-}$ $\rightarrow$ Cu
*Now, let us discuss the part (b), the displacement of Cu by Fe from copper sulphate solution.
*As in the electrochemical series iron is above the copper; it means that iron is more electronegative than the copper. The more reactive element displaces the less reactive metal.
*Thus, in the end we can conclude that copper will reduce in part (a), and in part (b) the reason of displacement is higher reactivity of iron than the copper.

Note: Don’t get confused while defining the reason for displacement. It is totally based on the concept of electrochemical reactivity series. Just remember the sequence of metals in the series. The oxidizing agent is for the element which reduces, and the reducing agent is for the element that oxidizes.