1.

Calculate the emf of the following cell at 298K, Fe(s)Fe^(2+)(0.001M)||H^+(1M)|H_2(g)(1ba r), Pt(s)

Answer»

Solution :The cell REACTION:
`Fe(s)+2H^+(aq)to Fe^(2+)(aq)+H_2(g) `
`E_(cell)^@=0.44V`
NERNST equation
`E_(cell)=E_(cell)^@-0.059/2log""([Fe^(2+)])/([H^+]^2)`
`E_(cell)=0.44V-0.059/2log""((0.001M))/(1M)^2`
`=0.44V-0.059/2log(10^-3)`
`=0.44V+0.0885 V`
`=0.5285V`


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