Write the cell reaction and calculate the e.m.f. of the following cell at 298K. Sn(s)|Sn2+(0.004M)||H+(0.020M)|H2(g)(1bar)|Pt(s). (Given: EoSn2+/Sn = -0.14V)

Oxidation at anode: Sn(s)→Sn2+(aq)+2e-
Reduction at cathode: 2H+(aq)+2e-→H2(g)
The cell reaction is
Sn(s)+2H+(aq)→Sn2+(aq)+H2(g)
Eocell=EoSHE-EoSn|Sn2+
Eocell=0.0V-(-0.14V)=0.14V
Ecell=Eocell - 0.0592/n log[Sn2+]×PH2/[H+]2
Ecell=0.14V - 0.0592/2 log(0.004M × 0.987atm)/[0.020]2
Note: 1 bar=0.987atm.
Ecell=0.111V