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The electrode potentials for

Cu^{2+}_{(aq)}+e^{-}\rightarrow Cu^{+}_{(aq)}

and Cu^{+}_{(aq)}+e^{-}\rightarrow Cu_{(s)}

are +0.15 V and +0.50, respectively. The value of E° Cu2+/ Cu will be:

  • Option 1)

    0.500 V

  • Option 2)

    0.325 V

  • Option 3)

    0.650 V

  • Option 4)

    0.150 V

 

Answers (1)

 

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

 

 Cu^{2+} (aq) +e^{-}\longrightarrowCu^{+}(aq), E_{1}^\circ =0.15 V

CU^{+}(aq)+e^{-}\longrightarrow Cu, E_{2}^\circ =0.50 V

Now \Delta G^\circ = \Delta G_{1}^\circ+\Delta G_{2}^{\circ}

-nFE^\circ =-n_{1}FE_{1}^\circ-n_{2}FE_{2}^\circ

=> E^\circ =\frac{n_{1}E_{1}^\circ+n_{2}E_{2}^\circ}{n}=\: \frac{1\times.15+1\times.50}{2}

=0.325V


Option 1)

0.500 V

THis is incorrect option

Option 2)

0.325 V

THis is correct option

Option 3)

0.650 V

THis is incorrect option

Option 4)

0.150 V

THis is incorrect option

Posted by

Vakul

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