
pH = -log10[H+
]
∴[H+ ]=10-pH
=100
=1
[H+] = 1M
The solution is strongly acidic
Zn(s) → Zn2+ (aq) + 2e-
Cu2+ (aq)+2e-
→ Cu(s)
Zn(s) +Cu2+ (aq) → Zn2+
(aq) + Cu(s)
Ecell =Eºcell – (0.0591/2) log ( [zn2+]/
[Cu2+] )
E1 =Eºcell - 0.0591/2 log (10-2/1)
E1 =Eocell
+ 0.0591 ........(1)
E2 =Eºcell - 0.0591/2 log (1/10-2)
E2 =Eocell
- 0.0591 .........(2)
E1>E2

Product
‘P’ in the above reaction is