1.

Calculate the emf of the following cell at 298 K: Cr(s)//Cr^(3+) (0.1 m)//Fe^(2+) (0.01 M)//Fe(s) [Given: E_("cell")^(@) = +0.30 V]

Answer»

Solution :The REACTION is:
`2Cr(s) + 3Fe^(2+)(s) (aq) to 3Fe(s) + 2Cr^(3+) (aq)`
In this case number of electrons involved, n=6
Apply the equation,
`E_("CELL") =E_("cell")^(@) -(2.303RT)/(NF) log ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
Substituting the values in the above equation, we have
`E_("cell") =0.30 -0.059/6 log [10^(-1)]^(2)/[10^(-3)]^(3) = 0.30 -0.059/6 log 10^(4) = 0.30 -0.04 =0.26 V`


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