Question

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.)

a. 1.0 M KF solution

b. 1.0 M CuCl₂ solution

c. 1.0 M Mgl₂ solution

Short answer

The reactions are as follows

(a) Cathode: $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

Anode:$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

(b) Cathode: ${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$

Anode:$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$.

(c) Cathode: $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

Anode: $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

Step-by-step solution

Conceptual Introduction

The electrode into which electricity flows is called the anode. The electrode where electricity is discharged or flowing out is called the cathode. Typically, the positive side is the anode.

Step 2(a): Reaction when it is electrolyzed with 1.0 M KF solution

Reaction of reduction for potassium can be denoted as,

${\mathrm{K}}^{+}+{\mathrm{e}}^{-}\to \mathrm{K}$

Reduction potential is -2.92 V.

Reaction of reduction for hydrogen gas can be denoted as,

$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

Reduction potential is -0.83 V.

The water reduction for hydrogen gas production has higher reduction potential and the reaction at cathode can be,

$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

The oxidation reaction for fluorine and water to produce oxygen gas can be denoted as,

$2{\mathrm{F}}^{-}\to {\mathrm{F}}_2+2{\mathrm{e}}^{-}$

reduction potential 2.87V

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

reduction potential -1.23V

The water reduction for hydrogen gas production has lower reduction potential and the reaction at anode will be,

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

Step 3(b) : Reaction when it is electrolyzed with 1.0 M CuCl2 solution

Reaction of reduction for copper can be denoted as,

${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$

Reduction potential is 0.34 V.

Reaction of reduction for hydrogen gas can be denoted as,

$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

Reduction potential is -0.83 V.

The copper reduction has higher reduction potential and the reaction at cathode can be,

${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$

The oxidation reaction for fluorine and water to produce oxygen gas can be denoted as,

${\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}$

reduction potential 1.36 V

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

reduction potential -1.23V

The chlorine oxidation reduction has higher reduction potential, so the reaction at anode will be,

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

Step 4(c): Reaction when it is electrolyzed with 1.0 M Mgl2 solution

Reaction of reduction for magnesium can be denoted as,

${\mathrm{Mg}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mg}$

Reduction potential is -2.37 V.

Reaction of reduction for hydrogen gas can be denoted as,

$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

Reduction potential is -0.83 V.

The water reduction for hydrogen gas production has higher reduction potential and the reaction at cathode can be,

$2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}$

The oxidation reaction for fluorine and water to produce oxygen gas can be denoted as,

${\mathrm{I}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{I}}^{-}$

reduction potential 0.54V

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

reduction potential -1.23V

The oxidation reaction for fluorine has higher reduction potential and the reaction at anode will be,

$2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$