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The cell reaction for the given cell is spontaneous of Pt(H_2)|H^+(1M)||H+(1M)|Pt(H_2)

  • Option 1)

    P_1>P_2

  • Option 2)

    P_1<P_2

  • Option 3)

    P_1=P_2

  • Option 4)

    P_1=1atm

 

Answers (1)

best_answer

As we learnt in 

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

 

 By Nernst eq.,we have

E_{({H_{2,P1}/{H_{2,P2}}})}=E \degree-\frac{RT}{nF}ln\, \frac{P_2}{P_1}

E \degree=0 for a concentration all

\Delta E_{cell}=\frac{-RT}{nF}log\left ( \frac{P2}{P1} \right )

So, since E should be positive for a spontaneous reaction, P_2<P_1

\therefore Solution\, \, is\, \, 1


Option 1)

P_1>P_2

Correct

Option 2)

P_1<P_2

Incorrect

Option 3)

P_1=P_2

Incorrect

Option 4)

P_1=1atm

Incorrect

Posted by

divya.saini

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