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  • Can someone explain - Electrochemistry - NEET-2

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 of CuSO4 changed to 0.01 M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2? (Given, \frac{{\text{RT}}} {\text{F}} = 0.059)

  • Option 1)

    E1 = E2

  • Option 2)

    E1 < E2

  • Option 3)

    E1 > E2

  • Option 4)

    \text{E}_2 = 0 \ne \text{E}_1

 

Answers (1)

best_answer

 

Electrode Potential(Nerst Equation) -

M^{n+}(aq)+ne^{-}\rightarrow M(s)

- wherein

E_{(M^{n+}/M)}=E_{(M^{n+}/M)}^{0}-\frac{RT}{nf}ln\frac{[M]}{[M^{n+}]}

[M^{n+}] is concentration of species

F= 96487 C moi^{-1}

R= 8.314 JK^{-1}moi^{-1}

T= Temperature in kelvin

 

 Cell reaction can be represented as

Zn + Cu2+ \rightarrow Cu + Zn2+

Applying in both cases

E^{\circ}=\frac{-0.0591}{2}.log\frac{Zn^{2+}}{Cu^{2+}}


Option 1)

E1 = E2

this is the incorrect option

Option 2)

E1 < E2

this is the incorrect option

Option 3)

E1 > E2

this is the correct option

Option 4)

\text{E}_2 = 0 \ne \text{E}_1

this is the incorrect option

Posted by

Plabita

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