JEE Main 10 Jan 2019 Shift 2 Solved Paper

Pt(s)|H2 (g, 1bar|HCl(aq)|Ag(s)|Pt(s)
10−6 m
Anod : H2 → 2H+ + 2e × 1
Cathode : e− + AgCl (s) → Ag (s) + Cl− (aq)
H2 (g)I + AgCl (s) → 2H+ + 2Ag(s) + 2Cl− (aq)
Ecell = E°cell -

0.06

2

log10((H+)2.(Cl−)2)
0.925 = (E°H2∕H+ + E°AgCl∕Ag.Cl− -

0.06

2

log10 ((10−6)2(10−6)2)
0.92 = 0 + E°AgCl∕Ag,Cl− - 0.03 log10(10−6)4
E°AgCl∕Ag,Cl− = 0.92 + 0.3 × - 24 = 0.2 V