emf of the following cell at 298 K in V is x ×10−2
Zn(s)|Zn2+(0.1M)||Ag+(0.01M)|Ag(s)
The value of x is____. (Rounded off to the nearest integer)
Given: E0Zn2+/Zn=−0.76V;E0Ag+/Ag=+0.80V;2.303RTF=0.059
Zn(s)|Zn+2(0.1M)||Ag+(0.01M)|Ag(s)
Zn(s)+2Ag+⇌2Ag(s)+Zn+2
E0=0.80+0.76=1.56;
Q=Zn2+(Ag+)2
E=E0−0.059nlog(Q)
E=1.56−0.0592log[0.1(0.01)2]
E=1.56−0.0592log[(10)3]
E=1.4715=147.15×10−2volt=x×10−2
x=147.15≈147