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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

In the electrochemical cell: Zn|ZnSO4 (0.01M)||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 - Chemistry

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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?

पर्याय

  • E1 < E2

  • E1 > E2

  • E2 ≥ E1

  • E1 = E2

MCQ
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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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पाठ 9: Electro Chemistry - Evaluation [पृष्ठ ६५]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
पाठ 9 Electro Chemistry
Evaluation | Q 20. | पृष्ठ ६५
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