Q

# Give answer! - Electrochemistry - NEET

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

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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

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