InterviewSolution
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InterviewSolution
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Determine of `pH`: The following cell has a potential of `0.55 V` at `25^(@)C`: `Pt(s)|H_(2)(1 bar)|H^(+)(aq. ? M)||Cl^(-)(1 M)|Hg_(2)Cl_(2)(s)|Hg(l)` What is the `pH` of the solution in the anode compartment? Strategy: First, read the shorthand notation to obtain the cell reaction. Then, calculate the half cell potential for the hydrogen electrode from the observed cell potential and the half cell poten-tial for the calomel reference electrode. Finally, apply the Nernst equation to find the `pH`. |
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Answer» The cell reaction is `H_(2)(g)+Hg_(2)Cl_(2)(s)hArr 2H^(+)(aq.)+2Hg(l)+2Cl^(-)(aq.)` and the potential is `E_("cell") = E_("anode") + E_("cathode")` `E_("cell") = E_(H_(2)//H^(+)) + E_(Hg_(2)Cl_(2)//Hg = 0.55 V` Because the reference electrode is the standard calomel electode, whic has `E = E^(@) = 0.28 V`, the half-cell potential for the hydrogn electode is `0.27 V`: `E_(H_(2)//H^(+)) = E_("cell") - E_(Hg_(2)Cl_(2)//Hg` `= (0.55 V) - (0.28 V)` `= 0.27 V` Applying the Nernst equation to the half-reaction `H_(2)(g) rarr 2H^(+)(aq.) + 2e^(-)`: `E_(H_(2)//H^(+)) = E_(H_(2)//H^(+))^(@) - ((0.0592 V)/(n)) "log" ((C_(H^(+))^(2))/(P_(H_(2))))` Substituting in the values of `E, E^(@)`, n and `P_(H_(2))` gives `0.27 V = (0V) - ((0.0592 V)/(2)) "log"((C_(H+)^(2))/(1))` `= (0.0592 V) (pH)` Therefore, the `pH` is `pH = (0.27 V)/(0.0592 V) = 4.6` |
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