Question

On the basis of standard electrode potential values, suggest which of the following reactions would take place?
(Consult the book for \(E^{\ominus}\)value).

Answer 1

For a reaction to take place spontaneously, value of \(E_{cell}^{\ominus}=\)positive
Because if \(E_{cell}^{\ominus}=\)positive, then \(\triangle G=\)negative

Since \(\triangle G=-nFE_{cell}^{\ominus}\)

And when \(\triangle G=\)negative, the reaction is spontaneous and reaction becomes feasible.

As we know that (from NCERT)

\(E_{Cu^{2+}/Cu}^{\ominus}=0.34V\)

\(E_{Zn^{2+}/Zn}^{\ominus}=-0.76V\)

\(E_{Mg^{2+}/Mg}^{\ominus}=-2.37V\)

\(E_{Fe^{2+}/Fe}^{\ominus}=-0.44V\)

\(E_{Br^{2+}/Br^{-}}^{\ominus}=1.08V\)

\(E_{Cl^{2+}/Cl^{-}}^{\ominus}=+1.36V\)

\(E_{Cd^{2+}/Cd}^{\ominus}=-0.76V\)

\(Cu+Zn^{2+}\rightarrow Cu^{2+}+Zn\)

In the given cell reaction, \(Cu\) is oxidised to \(Cu^{2+}\)
therefore \(Cu^{2+}/Cu\) couple acts as anode and \(Zn^{2+}\) is reduced to \(Zn\), therefore, \(Zn^{2+}/Zn\) couple acts as a cathode.

Half-cell reactions:

Oxidation: \(Cu\rightarrow Cu^{2+}+2e^{-};E^{\ominus}=0.34V\)
Reduction:\(\underline{Zn^{2+}+2e^{-}\rightarrow Zn;E^{\ominus}=-0.76V}\)
\(Cu+Zn^{2+}\rightarrow Cu^{2+}+Zn; E_{cell}^{\ominus}=-1.1V\)

Since value of \(E_{cell}^{\ominus}=\)negative, this reaction is not feasible and does not take place.
B. \(Mg+Fe^{2+}\rightarrow Mg^{2+}+Fe\)

In the given cell reaction, Mg is oxidised to \(Mg^{2+}\)
, therefore \(Mg^{2+}/Mg\) couples acts as anode and \(Fe^{2+}\)
is reduced to Fe , therefore, \(Fe^{2+}/Fe\) couple acts as a cathode.
Half-cell reactions:

Oxidation: \(Mg\rightarrow Mg^{2+}+2e^{-};E^{\ominus}=-2.36V\)
Reduction:\(\underline{Fe^{2+}+2e^{-}\rightarrow Fe;E^{\ominus}=-0.44V}\)
\(Mg+Fe^{2+}\rightarrow Mg^{2+}+Fe; E_{cell}^{\ominus}=1.92V\)
Since value of \(E_{cell}^{\ominus}=\)positive, this reaction is feasible, thus Mg will be oxidised on losing electrons and \(Fe^{2+}\)will be reduced on gaining electrons.

C.\(Br_{2}+2Cl^{-}\rightarrow Cl_{2}+2Br^{-}\)

In the given cell reaction, \(Cl^{-}\) is oxidised to \(Cl_{2}\), therefore \(Cl_{2}/Cl\)couple acts as anode and \(Br_{2}\) is reduced to \(Br_{2}/Br^{-}\) couple acts as a cathode.

Oxidation: \(Br^{2+}+2e^{-}\rightarrow Br;E^{\ominus}=1.09V\)
Reduction:\(\underline{2Cl^{-}\rightarrow Cl_{2}+2e^{-};E^{\ominus}=1.36V}\)

\(Br^{2}+Cl^{-}\rightarrow +2Br^{-} ;E_{cell}^{\ominus}=-0.27V\)

Since value of \(E_{cell}^{\ominus}=\)negative, this reaction is feasible.

D.\(Fe+Cd^{2+}\rightarrow Cd +Fe^{2+}\)

In the given cell reaction, \(Fe\) is oxidised to \(Fe^{2+}, \)
therefore \(Fe^{2+}/Fe\) couple acts as anode and \(Cd^{2+}\) is reduced to \(Cd\), therefore , \(Cd^{2+}/Cd\) the couple acts as a cathode.

Half-cell reactions:
Oxidation: \(Fe\rightarrow Fe^{2+}+2e^{-};E^{\ominus}=-0.44 V\)
Reduction:\(\underline{Cd^{2}+2e^{-}\rightarrow Cd;E^{\ominus} =0.40V}\)

\(Fe+Cd^{2}\rightarrow Cd^{2+}+Fe^{2+} ;E_{cell}^{\ominus}=0.04V\)

Since value of \(E_{cell}^{\ominus}=\) positive, this reaction is feasible, thus \(Fe\) will be oxidised on losing electron and \(Cd^{2+}\) will be reduced on gaining electrons.

Thus, based on standard electrode potential values, reaction in (B) and (D) can takes place as in both reaction \(E_{cell}^{\ominus}\)
is positive and thus reaction is spontaneous