Question

The two half cell reactions and their oxidation potentials are

1. \[\ce{Pb_{(s)} - 2e- -> Pb{^{2+}_{(aq)}}, E^{\circ}_{Pb/Pb^{2+}}}\] = +0.13 V
2. \[\ce{Ag_{(s)} - e- -> Ag{^+_{(aq)}}, E^{\circ}_{Ag/Ag^{+}}}\] = −0.80 V

Write the cell reaction and calculate the cell emf.

Answer 1

Given:

1. \[\ce{Pb_{(s)} - 2e- -> Pb{^{2+}_{(aq)}}}\],
2. \[\ce{Ag_{(s)} - e- -> Ag{^+_{(aq)}}}\],

Oxidation potential of Pb is Pb²⁺; \[\ce{E{^{\circ}_{cell}}}\] = +0.13 V

Oxidation potential of Ag is Ag⁺; \[\ce{E{^{\circ}_{cell}}}\] = −0.80 V

Higher oxidation potential is more likely to oxidise.
Here, Pb has a higher oxidation potential (+0.13 V), so Pb gets oxidised to the anode. Ag gets reduced to cathode.

Anode (oxidation): \[\ce{Pb_{(s)} -> Pb{^{2+}_{(aq)}} + 2e-}\]

Cathode (reduction): \[\ce{2Ag{^+_{(aq)}} + 2e- -> 2Ag_{(s)}}\]

Overall cell reaction: \[\ce{Pb_{(s)} + 2Ag{^+_{(aq)}} -> Pb{^{2+}_{(aq)}} + 2Ag_{(s)}}\]

Convert oxidation potential to reduction potentials

\[\ce{E_{Pb^{2+}/Pb}}\] = −0.13 V

\[\ce{E_{Ag^{+}/Ag}}\] = +0.80 V

We know that

\[\ce{E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}}\]

= 0.80 − (−0.13)

= 0.93 V