Predicating the spontaneity of cell reaction: Predict whether following reaction would spontaneously as written at `298 K`:
`Co(s) + Fe^(2+)(aq.) rarr Co^(2+)(aq.) + Fe(s)`
given that `C_(Co^(2+)) = 0.15 M` and `C_(Fe^(2+)) = 0.68 M`
Strategy: Calculate `E_("cell")` by applying the Nernst equaltion to the overall cell reaction. The cell is spontaneous in the direction written, for the concentration given, if `E_("cell")` is positive. However the reverse reaction would be favoured at those concentrations, if resulting cell potential is negative.

Explanation

The half-reactions are
`{:("Oxidation (Anode)": Co(s)hArr CO^(2+)(aq.)+2e^(-)),("Reduction (Cathode)": Fe^(2+)(aq.)+2e^(-)hArr Fe(s)):}`
Using Table 3.1 we find that
`E_(Co^(2+)//Co)^(@) = -0.28 V and E_(Fe^(2+)//Fe)^(@) = -0.44 V`
Therefore, the standard emf of the cell is
`E_("cell")^(@) = E_(cathode)^(@) - E_(Anode)^(@)`
`= (-0.44V) - (-0.28V)`
`= -0.16V`
Using Nernst equation, we write
`E_("cell") = E_("cell")^(@) - (0.0592V)/(n) "log Q"`
`E_("cell")^(@) - (0.0592V)/(n) "log" C_(Co^(2+))/(C_Fe^(2+))`
`= -0.16 V - (0.0592 V)/(2) "log" (0.15)/(0.68)`
`= -0.16 V + 0.019 V`
`= -014 V`
Since `E_("cell")` is neagative (or `DeltaG_("cell")` is positive), the ration is non-spontaneous in the direction written.