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प्रश्न
In the electrochemical cell: Zn|ZnSO4 (0.01 M)||CuSO4 (1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that CuSO4 changed to 0.01 M, the emf changes to E2. From the above, which one is the relationship between E1 and E2?
विकल्प
E1 < E2
E1 > E2
E2 ≥ E1
E1 = E2
उत्तर
E1 > E2
Explanation:
`"E"_"cell" = "E"_"cell"^0 - 0.0591/2 log (["Zn"^(2+)])/(["Cu"^(2+)])`
E1 = `"E"_"cell"^0 - 0.0591/2 log 10^-2/1`
E1 = `"E"_"cell"^0 + 0.0591` .....................(1)
\[\ce{Zn_{(s)} -> Zn^{2+}_{( aq)} + 2e^-}\]
\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]
\[\ce{Zn_{(s)} + Cu^{2+}_{( aq)} -> Zn^{2+}_{( aq)} + Cu_{(s)}}\]
E2 = `"E"_"cell"^0 - 0.0591/2 log 1/10^-2`
E2 = `"E"_"cell"^0 - 0.0591` .....................(2)
∴ E1 > E2
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