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 Nerst, `E = E^(@) - (0.059)/(2) "log" ([Zn^(2+)])/([Fe^(2+)])`
`E^(@)= -0.44 + 0.763`
=0.323V
`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