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Discuss the method to determine cell potential of any cell when standard hydrogen electrode is considered as cathode with suitable example. |
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Answer» Solution :At 298K the EMF of the cell, standard hydrogen electrode and second half-cell constructed by taking standard hydrogen electrode as cathode (reference half-cell) and the other half-cell as anode, gives the reduction potential of the other half-cell. Electrode of the other half-cell as anode || Standard hydrogen electrode as cathode E.g.,general cell : `M_((S))|M_((aq))^(n+)(1M)||H_((aq))^(+)(1M)|(1)/(2)|(1)/(2)H_(2(g))(1" bar")|Pt` and `E_(cell)^(THETA)=(E_(R)^(Theta)-E_(L)^(Theta))` but `E_(R)^(Theta)=0.0V` `E_(cell)^(Theta)=-E_(L)^(Theta)` (for such cel the cell potential is negative) e.g., anode of zinc half-cell || cathode of hydrogen electrode half-cell `Zn_((S))|Zn_((aq))^(2+)(1M)||H_((aq))^(+)(1M)|H_(2(g))(1" bar")|Pt_((S))` For above cell, practical value of reduction potential is `-0.76V`. So, for `Zn_((aq))^(2+)(1M)+2e^(-)toZn_((S))`, standard reduction potential `=0.76V` i.e., `E_(Zn^(2+)|Zn)^(Theta)=-0.76V` NOTE: When standard hydrogen electrode is on right side and working as cathode then other half-cell has negative reduction potential value and such cell possess negative emf value. Reduction reaction occurs on cathode electrode present on right side of cell: Reaction: `H_("aq, 1M")^(+)+e^(-) to (1)/(2)H_(2(g))(1" bar")` The negative value of such cell shows oxidation of zinc metal is done by `H^(+)` ions and zinc metal get dissolved in HCl acid. also `H^(+)` ions get reduced by zinc metal. `Zn_((S))+2H_((aq))^(+) to Zn_((aq))^(2+)+H_(2(g))` |
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