1.

What is the correct Nernst formula of calculate potential of reaction Zn_((aq))^(2+)+2e^(-) to Zn_((S))

Answer»

`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(THETA)-(0.059)/(2)LOG[Zn_((aq))^(2+)]`
`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(Theta)-(0.059)/(2)log"(1)/([Zn_((aq))^(2+)])`
`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(Theta)+(0.059)/(2)log[Zn_((aq))^(2+)]`
`E_(Zn^(2+)|Zn)=E_(Zu^(2+)|Zn)^(Theta)+(0.059)/(2)"log"(1)/([Zn_((aq))^(2+)])`

Solution :`Zn_((aq))^(2+)+2E^(-) to Zn_((S))`
Where, Reactants=`[Zn_((aq))^(2+)],` Product=[Zn]
`therefore "log "K_(c)=(["Product"])/(["Reactant"])=([Zn_((S))])/([Zn_((aq))^(2+)])=(1)/([Zn^(2+)])`
`therefore E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(Theta)-(0.059)/(2)"log"(1)/([Zn_((aq))^(2+)])`
Or `E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(Theta)+(0.059)/(2)log[Zn^(2+)]`


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