1.

Zn|Zn^(2+)(a=0.1 M)||Fe^(2+)(a=0.01M)|Fe the emf of the above cell is 0.2905 V Equilibrium constant for thecellreactionis

Answer»

`10^(0.32//0.591)`
`10^(0.32//0.0295)`
`10^(0.26//0.295)`
`10^(0.32//0.295)`

Solution :Forcell `Zn|Zn^(2+)(a=0.1 M)||Fe^(2+) (a=0.01 M)|Fe`
The halfcell reactionare
`I Zn(s) rarr Zn^(2+) (aq)+2e^(-)`
`ii Fe^(2+) (Aq)+2e^(-) rarr Fe(s)`
`Zn(s) +Fe^(2+) (Aq) rarr Zn^(2+)(aq)+Fe(s)`
on applyingNernstequation
`E_(cell) =E_(cell)^(@) =0.2905 +0.295 =0.32 V`
At equilibrium `E_(cell)=0`
`E_(cell)^(@) =(0.0591) /(N ) log_(10) k_(c )`
`0.32 =(0.591)//(2) log_(10)k_(c )` or `k_(c )=10^(0.32//0.295)`


Discussion

No Comment Found

Related InterviewSolutions