Write Nernst equation and calculate the e.m.f of the following cell at 298 K:
`Zn|Zn^(2+)(0.01M)||Fe^(2+)(0.005M)|Fe`
Given that:
`(Zn^(2+)//Zn)=-0.763V and E_(Fe^(2+)//Fe)^(@)=-0.44V" "log2=0.3010`.

Explanation

At anode, `Zn rarr Zn^(2+) + 2e^(-)`
At cathode,
`ul(Fe^(2+) + 2e rarr Fe)`
`ul(Zn+Fe^(2+) rarr Zn^(2+) +Fe)`
According to Nernst, `E= E^(@)- (0.059)/(2) "log" ([Zn^(-1)])/([Fe^(2+)])`
`E+ 0.323 - (0.059)/(2) "log" (0.01)/(0.005)`
`=0.323 - (0.059)/(2) log 2`
`=0.323 - (0.059 xx 0.301)/(2)`
= 0.31V