1.

The standard oxidation ptoetial of zinic is 0.76 V and of silveris -0.80 V calculatethe emf of thecell Zn|Zn(NO_(3))_(2)(0.25 M)||AgNO_(3)(0.1)M|Ag "at" 25^(@) C

Answer»

1.158 V
1.352 V
0.768 V
21.32 V

Solution :Thecell REACTIONIS `Zn+ 2Ag^(+) RARR 2 Ag+ Zn^(2+)`
`E_("cell")^(@)=E_("oxi")^(@)` of `Zn + e_("red")^(@)` of`Ag=0.76 +0.80 =1.56 V`
We KNOW that
`E_("cell") =E_("cell")^(@)-(0.0591)/(n) log[("products")]/[("reactants")] =E_("cell")^(@) -(0.0591)/(2) log(0.25)/(0.1xx0.1)`
`=1.56 -(0.0591)/(2) xx (1.0591)/(2)xx1.3979 =1.3979 =(1.56 -0 0.0413) V, E_("cell")=1.5187 V`


Discussion

No Comment Found

Related InterviewSolutions