KEY2CHEM

Electrochemical Reactions

Electrochemistry is the study of the relationship between chemical change and electrical work. The type of chemical reaction that occurs via a net movement of electrons is an oxidation-reduction (commonly called redox) reaction. A spontaneous redox reaction is one that occurs with a positive cell (or half-cell) potential, since the relationship between cell potential and free energy change is \(\Delta{G}^\circ = -nFE^{\circ}\), where \(F =\) Faraday’s constant (\(96,485\frac{\text{C}}{\text{mol e}^-}\)). 


Example 1.

Which half reaction is spontaneous?

A. \(\require{mhchem}\ce{Zn^2+(aq) + 2 e- -> Zn(s) \;\;\;\;E^{\circ}} = -0.76 \;V\)

B. \(\require{mhchem}\ce{Cu^2+(aq) + 2 e- -> Cu(s) \;\;\;\;E^{\circ}} = 0.34 \;V\)

C. \(\require{mhchem}\ce{Ni^2+(aq) + 2 e- -> Ni(s) \;\;\;\;E^{\circ}} = -0.25 \;V\)

 

 

 

Solution

B. \(\require{mhchem}\ce{Cu^2+(aq) + 2 e- -> Cu(s) \;\;\;\;E^{\circ}} = 0.34 \;V\)

A positive value of \(E^\circ\) corresponds to a negative value of \(\Delta G^\circ\), and the reaction is spontaneous. 


Example 2.

A galvanic electrochemical cell contains the following redox reaction. Which species is the anode in this galvanic cell?

\(\require{mhchem}\ce{2 Ag+(aq) + Fe(s) -> 2 Ag(s) + Fe^2+(aq)}\)

 

A. \(\require{mhchem}\ce{Ag+(aq)}\)

B. \(\require{mhchem}\ce{Fe(s)}\)

C. \(\require{mhchem}\ce{Ag(s)}\)

 

 

 

 

 

Solution

B. \(\require{mhchem}\ce{Fe(s)}\)

Oxidation occurs at the anode in an electrochemical cell. Since \(\require{mhchem}\ce{Fe(s)}\) is undergoing oxidation, \(\require{mhchem}\ce{Fe(s)}\) is the anode. \(\require{mhchem}\ce{Ag(s)}\) is the cathode, since \(\require{mhchem}\ce{Ag+(aq)}\) is reduced to \(\require{mhchem}\ce{Ag(s)}\).


Example 3.

 

How many minutes will it take to electrolyze \(2.99\text{ g}\) of \(\require{mhchem}\ce{Ni(s)}\) from an aqueous solution of \(\require{mhchem}\ce{NiSO4}\) if a constant current of \(15.1\text{ A}\) is applied?

A. \(10.9\text{ minutes}\)

B. \(5.43\text{ minutes}\)

C. \(21.7\text{ minutes}\)

 

 

 

 

 

Solution

A. \(10.9\text{ minutes}\) 

This is a stoichiometry problem that can be solved using Faraday’s laws.

\(\require{mhchem}\ce{Ni^2+(aq) + 2 e- -> Ni(s)}\)

\(2.99\text{ g}\times \frac{1\text{ mol Ni}}{58.69\text{ g Ni}} \times \frac{2\text{ mol e}^-}{1\text{ mol Ni}}\times \frac{96,485\text{ C}}{1\text{ mol e}^-}\times \frac{1 \text{ second}}{15.1\text{ C}}\times \frac{1\text{ minute}}{60\text{ seconds}} = 10.9\text{ minutes}\)