Question

2. Consider the following half-cell reactions and associated standard half-cell potentials and
determine the maximum voltage that can be obtained by combination resulting in
spontaneous processes:
AuBră (aq) + 3e -Au(s) + 4Br+ (aq); EⓇ=-0.86 V
Eu 3+ (aq) + e- Eu2+ (aq); E°-0.43V
Sn2+ (aq) + 2e- Sn(s); E°=-0.14 V
10+ (aq) + H20(1+ 2e r (aq) + 20H(aq); E° = +0.49 V
(a) +0.72
(b) +1.54 - (c) +1.00
(d) +1.35​

Answer 1

Answer: The best reducing agent is the species which can be easily oxidized. In other words, the best reducing agent is the species which has the highest oxidation potential.

In this half-cell reaction:

PO43- (aq) + 2 H2O (l) + 2 e- --> HPO32-(aq) + 3 OH- (aq)

Eo= -1.05 V

P = +5 in PO43-

P = +3 in HPO32-

Therefore this is a reduction half-equation. To obtain the oxidation half-equation we have to reverse it;

HPO32-(aq) + 3 OH- (aq) --> PO43- (aq) + 2 H2O (l) + 2 e-

Eo= +1.05 V

Similarly;

In this half-cell reaction:

PbO2 (s) + H2O (l) + 2 e- --> PbO (s) + 2 OH- (aq)

Eo= +0.28 V

Pb = +4 in PbO2

Pb = +2 in PbO

This is also a reduction half-equation. The oxidation half-equation:

PbO (s) + 2 OH- (aq) ---> PbO2 (s) + H2O (l) + 2 e-

Eo= - 0.28 V

In this half-cell reaction:

IO3- (aq) + 2 H2O (l) + 4 e- --> IO- (aq) + 4 OH- (aq)

Eo= +0.56 V

I = +5 in IO3-

I = +1 in IO-

The oxidation half-equation:

IO- (aq) + 4 OH- (aq) ---> IO3- (aq) + 2 H2O (l) + 4 e-

Eo= -0.56 V

Since the oxidation potential of

HPO32-(aq) + 3 OH- (aq) --> PO43- (aq) + 2 H2O (l) + 2 e-

Eo= +1.05 V

is the highest (largest), the species which is oxidized in this equation ( HPO3^2- ) is the best reducing agent.