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Calculate the Zn-Cu cell (Daniell) potential. |
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Answer» Solution :`(E_(Zn^(2+)|Zn)^(THETA)=-0.76V and E_(Cu^(2+)|Cu)^(Theta)=0.34V)` According to the convention, anode half-cell has less REDUCTION potential and CATHODE half-cell has high reduction potential so, oxidation reaction on the zinc electrode is available on left side of cell is as follows: `Zn_((S))toZn_((aq," 1M")^(2+)+2e^(-)`. . . (i) Reduction reaction on the copper electrode available on right side of cell is as follows: `Cu_((aq," 1M"))^(2+) +2e^(-) to Cu_((S))` . . . (ii) Sum of reaction (i) and (ii) gives redox reaction which is as follows: `Zn_((S))+Cu_((aq," 1M"))^(2+)=Zn_((aq," 1M"))^(2+)+Cu_((S))` Now emf of this cell `=E_(cell)^(Theta)=E_(R)^(Theta)E_(L)^(Theta)` `THEREFORE DeltaE_(cell)^(Theta)=-0.34V-(-0.76V)=1.10V` |
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