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Given:

  1. Cu^{2+}+2e^{-}\rightarrow Cu, E^{\circ}=0.337 V
  2. Cu^{2+}+e^{-}\rightarrow Cu^{+}, E^{\circ}=0.153 V

Electrode potential, E° for the reaction,

Cu^{+}+e^{-}\rightarrow Cu^{+}, will \: be:

  

  • Option 1)

    0.90 V

  • Option 2)

    0.30 V

  • Option 3)

    0.38 V

  • Option 4)

    0.52 V

 

Answers (1)

best_answer

 

Gibbs energy of the reaction -

\Delta_{r}G=-nFE_{cell}

- wherein

\Delta _{r}G = gibbs energy of the reaction

E_{cell}= emf of the cell

nF = amount of charge passed

 

 Given,

cu^{2+}+2e^{-}\rightarrow cu \ \ E_{1}^{0}=0.337V

cu^{2+}+2e^{-}\rightarrow cu^{+} \ \ E_{2}^{0}=0.153V

Required reaction is

cu^{+}+e^{-}\rightarrow cu, \ E^{3}^{0}=?

Applying \Delta G_{3}^{0}=-nFE^{0}

\Delta G_{3}^{0}=\Delta G_{1}^{0}-G_{2}^{0}

-nFE_{3}^{0}=-n_{1}FE_{1}^{0}-\left ( -n_{2}FE_{2}^{0} \right )

E_{3}^{0}=2E_{1}^{0}-E_{2}^{0}

=2\times .337-0.153=0.52V


Option 1)

0.90 V

This is incorrect option

Option 2)

0.30 V

This is incorrect option

Option 3)

0.38 V

This is incorrect option

Option 4)

0.52 V

This is correct option

Posted by

Aadil

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