Question

Zn2+ (aq) + 2 e- → Zn (s)

Cu2+ (aq) + 2 e-Cu (s)

a. Are these oxidation or reduction half reactions?

b. Calculate ΔG° for each of these reactions. Note that ΔG°f for the electron is 0 kJ/mol.

c. Based on your answer to (b), does Zn2+ or Cu2+ more strongly favor reduction?

Answer 1

Zn²⁺ (aq) + 2 e^- → Zn(s)

Cu²⁺ (aq) + 2 e^-→Cu(s)

(a)The above reactions are reduction half reactions as electrons are gained during these half reactions.

(b)The standard reduction potential for the Zn²⁺ /Zn is given by E^o_{(Zn²⁺ /Zn)}=-0.76V

G^o = -nFE^o

Where,

n= no. of electrons gained during reduction of metal ion

F= Faradays constt =96485 C mol^-1

E^o= Standard reduction potential

For the reaction,

Zn²⁺ (aq) + 2 e^- → Zn(s)

G^o = -nFE^o_{(Zn²⁺ /Zn)}

=- 2 x 96485 C mol^-1 x -0.76V

=146,657.2 J/mol

=146.66 kJ/mol

For the reaction

Cu²⁺ (aq) + 2 e^-→Cu(s)

The standard reduction potential for the Cu²⁺ /Cu is given by E^o_{(Cu²⁺ /Cu)}= 0.34V

G^o = -nFE^o_{(Cu²⁺ /Cu)}

=- 2 x 96485 C mol^-1 x 0.34V

=-65,609.8 J/mol

= -65.61 kJ/mol

(c) Cu²⁺ strongly favours reduction as it has a negative G^o value for the reduction process, and therefore reduction of  Cu²⁺ is a spontaneous process

G^o value for the reduction of Zn²⁺ is positive, therefore reduction of  Zn²⁺ is a non-spontaneous process.