Question

The half cell reaction with their oxidation potentials are

Pb(s) → Pb²⁺(aq) + 2e^-; E°cell = + 0.13V

Ag(s) → Ag(aq) + e^-; E°cell = -0.80V

Write the cell reaction and calculate its EMF.

Answer 1

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