Q21.51 P - Chemistry: Molecular Nature Of Matter And Change

Question

When a clean iron nail is placed in an aqueous solution of copper(II) sulfate, the nail becomes coated with a brownish black material. 

(a) What is the material coating the iron? 

(b) What are the oxidizing and reducing agents?

 (c) Can this reaction be made into a voltaic cell? 

(d) Write the balanced equation for the reaction. 

(e) Calculate E°cell for the process.

Step-by-Step Solution

Answer

(a) Copper metal is coating the iron nail.

(b)The oxidizing agent is copper (II) sulfate, and the reducing agent is iron metal.

(c)This reaction can be made into a voltaic cell.

(d)  _{${\text{Cu}}_{\text{(aq)}}^{\text{2 + }}{\text{ + Fe}}_{\text{(s)}}\to {\text{Cu}}_{\text{(s)}}{\text{ + Fe}}^{\text{2 + }}$}

(e)  _{$E^{\circ }cell=0.78V$}

Step 1:Material coating over iron

When an iron nail is dipped into Copper(II) sulfate, iron is oxidized and forms Ferrous sulfate. The Cu metal formed deposits on the iron nail. The Cu metal deposits formed are brownish black in color.

 ${\text{Fe + CuSO}}_{\text{4}}\to {\text{FeSO}}_{\text{4}}\text{ + Cu}$

Step 2: Oxidizing and reducing agents Let us write down the redox reaction to identify the oxidizing and reducing agents.

$\stackrel{0}{\mathrm{Fe}}+\stackrel{+2}{\mathrm{Cu}}\stackrel{-2}{{\mathrm{SO}}_4}\to \stackrel{+2}{\mathrm{Fe}}{\stackrel{-2}{\mathrm{SO}}}_4+\stackrel{0}{\mathrm{Cu}}$

As it is evident from the change in oxidation numbers that Fe is oxidized and ${\text{Cu}}^{\text{ + 2}}$  is reduced. Therefore it can be said, that iron has reduced copper and copper has oxidized iron. Thus, Fe is a reducing agent while Cu is an oxidizing agent.

Step 3: Voltaic cell

The working principle of a voltaic cell depends on the spontaneity of the redox reaction taking place between the chemical species. If the redox reaction is spontaneous $∆G<0$ the potential difference between the two electrodes generates electrical energy.

In the above situation when an iron nail is dipped into copper sulfate solution, the copper metal deposits on iron nail spontaneously. Thus this reaction can be made into a voltaic cell.

Step 4: Voltaic cell The equation representing the voltaic cell is given below.

$$\begin{gathered} \underline{{\text{Cu}}^{\text{ + 2}}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }}\to \text{Cu}\left(\text{s}\right) \\ \text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{ + 2}}\left(\text{aq}\right){\text{ + 2e}}^{\text{ - }}} \\ {\text{Cu}}_{\text{(aq)}}^{\text{2 + }}{\text{ + Fe}}_{\text{(s)}}\to {\text{Cu}}_{\text{(s)}}{\text{ + Fe}}^{\text{2 + }}\left(\text{aq}\right) \end{gathered}$$

Step 5: E cell

The ${\text{E}}_{\text{cell}}$  can be calculated as the difference in the electrode potentials of the half cell.

 ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cathode}}{\text{ - E}}_{\text{anode}}$

Let us determine the anode and cathode of the cell. Oxidation occurs at anode and reduction occurs at cathode.

 $$\begin{gathered} \text{Anode} \text{reaction} {\text{Cu}}^{\text{ + 2}}{\text{ + 2e}}^{\text{ - }}\to \text{Cu} {\text{E}}^{\text{0}}\text{ = - 0.44} \text{V} \\ \text{Cathode} \text{reaction} \text{Fe}\to {\text{Fe}}^{\text{ + 2}}{\text{ + 2e}}^{\text{ - }} {\text{E}}^{\text{0}}\text{ = 0.38} \text{V} \end{gathered}$$

$\begin{array}{rcl}& & {{\text{E}}^{\text{0}}}_{\text{cell}}\text{ = }{{\text{E}}^{\text{0}}}_{\text{cathode}}\text{ - }{{\text{E}}^{\text{0}}}_{\text{anode}}\\ & & \text{ = 0.34} \text{V - }\left(\text{ - 0.44} \text{V}\right)\\ & & \text{ = 0.78} \text{V}\\ & & \end{array}$

Hence, the cell potential is  $0.78 \text{V}$