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The standard Gibbs energy for the given cell reaction in KJ\; mol^{-1} at 298\; K is :

Zn\left ( s \right )+Cu^{2+}\left ( aq \right )\rightarrow Zn^{2+}\left ( aq \right )+Cu\left ( s \right ),

E^{0}=2\; V\; at\; 298\; K

(Faraday's constant, F=96000\; C\; mol^{-1})

  • Option 1)

     - 384

  • Option 2)

     384

  • Option 3)

    192

  • Option 4)

    -192

 

Answers (1)

best_answer

We know, \Delta G^{\circ} =-nFE^{\circ}_{cell}=-2\times 96500\times 2.0

                 \Delta G^{\circ}=-386\times 10^{3}

                        \Delta G^{\circ} =-386\; Kj/mol

           


Option 1)

 - 384

Option 2)

 384

Option 3)

192

Option 4)

-192

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solutionqc

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