InterviewSolution
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InterviewSolution
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The Nearst Equation: Calcualte the nonstandard electrode potential, `E`, for the `Fe^(3+)//Fe^(2+)` electrode when the concentration of `Fe^(2+)` is exactly firve time that of `Fe^(3+)` Strategy: The Nearst equation helps us calculate electrode potentials for concentrations other than one molar. The Tabulation of standard electrode potentials gives us the value of `E^(@)` for the reduction half-reaction. Use the balanced half-reaction and the given concentration ratio to calculate the value of `Q`. Finally substitude this into the Nearst equation with `n` equal to the number of moles of elecrons involved in the half-reaction. |
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Answer» The reduction half-reaction is `Fe^(3+)(aq.) + e^(-) rarr Fe^(2+)(aq.)E_(Fe^(3+)//Fe^(2+)) = + 0.77V` Calculating the value of `Q` by using the fact that the concentration of `Fe^(2+)` is five times that of `Fe^(3+)`: `Q = ["Red"]^(b)/["Ox"]^(a) = C_(Fe^(2+))/C_(Fe^(3+)) = (5C_(Fe^(3+)))/(C_(Fe^(3+))) = 5` The balanced half-cell reaction shows one mole of electrons, or `n = 1`. Putting values into the Nearst equation, `E_(Fe^(3+)//Fe^(2+))^(@) = E_(Fe^(3+)//Fe^(2+))^(@) - (0.0592 V)/(n) log Q` `= +0.77 V - (0.0592 V)/(1) log 5` `= 0.77 V - 0.04 V` `= + 0.73` |
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