1.

Derive Nernst equation for the cell. Ni(s)|Ni^(2+) (aq. 0.1 M)||Ag^(+) (aq. 0.1 M)|Ag (s) and also find its cell potential. Given : E_(Ag^(+)//Ag)^(@)=0.80 volt and E_(Ni^(2+)//Ni)^(@)=-0.25 volt

Answer»

Solution :Cell reaction,
`NI+2Ag^(+) rarr 2Ag+Ni^(2+),""E_(cell)^(@)=1.05` volt
Nernst EQUATION, `E_(cell)=E_(cell)^(@)-0.0591/2 "log "([Ni^(2+)])/([Ag^(+)]^(2))`
`=(1.05-0.0295)` volt
`=1.02` volt


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