The half cell reaction with their oxidation potentials arePb(s) → Pb

1.	The half cell reaction with their oxidation potentials arePb(s) → Pb2+(aq) + 2e-; E°cell = + 0.13VAg(s) → Ag(aq) + e-; E°cell = -0.80VWrite the cell reaction and calculate its EMF.
Answer» Rewrite the two equations in the reduction form thus. Pb²⁺(aq) + 2eE^– → Pb(s); E° = -0.13V ...(i) Ag⁺(aq) + e^– → Ag(s); E° = + 0.80V …(ii) To obtain the equation for the cell reaction, multiply Eq. (ii) with 2 and subtract Eq. (i) from it, we have, Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s); E°cell = + 0.80 - (-0.13) = + 0.93V

The half cell reaction with their oxidation potentials arePb(s) → Pb2+(aq) + 2e-; E°cell = + 0.13VAg(s) → Ag(aq) + e-; E°cell = -0.80VWrite the cell reaction and calculate its EMF.

Answer»

Rewrite the two equations in the reduction form thus.

Pb²⁺(aq) + 2eE^– → Pb(s);

E° = -0.13V ...(i)

Ag⁺(aq) + e^– → Ag(s);

E° = + 0.80V …(ii)

To obtain the equation for the cell reaction, multiply Eq. (ii) with 2 and subtract Eq. (i) from it, we have,

Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s);

E°cell = + 0.80 - (-0.13)

= + 0.93V