For the disproportionation reaction 2 Cu ^{+} | vedclass

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Question

(B) The disproportionation reaction is:$2 Cu ^{+}( aq ) \longrightarrow Cu ( s ) + Cu ^{2+}( aq )$This can be split into two half-reactions:Oxidation:

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(B) The disproportionation reaction is:$2 Cu ^{+}( aq ) \longrightarrow Cu ( s ) + Cu ^{2+}( aq )$This can be split into two half-reactions:Oxidation: $Cu ^{+}( aq ) \longrightarrow Cu ^{2+}( aq ) + e ^{-} \quad (E^{0} = -0.16 \ V)$Reduction: $Cu ^{+}( aq ) + e ^{-} \longrightarrow Cu ( s ) \quad (E^{0} = 0.52 \ V)$$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 0.52 \ V - 0.16 \ V = 0.36 \ V$At equilibrium,the relationship between $E_{cell}^{0}$ and the equilibrium constant $K$ is given by:$E_{cell}^{0} = \frac{RT}{nF} \ln K$Here,$n = 1$ (number of electrons transferred).$\ln K = \frac{E_{cell}^{0} 	imes n}{RT/F} = \frac{0.36 	imes 1}{0.025} = 14.4$Expressing $14.4$ as $	imes 10^{-1}$:$14.4 = 144 	imes 10^{-1}$Thus,the value is $144$.

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