What is the emf of a cell consisting of a Pb2 / Pb half-cell and a Pt / H / H2 half-cell if [Pb2 ]

0.083 V

Explanation:

The equation of the reaction is;

Pb(s) + 2H^+(aq) > Pb^2+(aq) + H2(g)

E°cell = E°cathode - E°anode

E°cathode = 0.00 V

E°anode = -0.13 V

E°cell = 0.00-(-0.13)

E°cell = 0.13 V

Q= [Pb^2+] pH2/[H^+]^2

Q= 0.1 × 1/[0.05]^2

Q= 0.1/0.0025

Q= 40

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

Ecell= 0.13 - 0.0592/2 log (40)

Ecell = 0.13 - 0.047

Ecell= 0.083 V

the assestment will probley apele with cause and effect

if they have a finite resultant