Cu^{+} ion is not stable in aqueous solution | vedclass

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(D) The disproportionation reaction is $2Cu^{+} 	o Cu^{2+} + Cu$.This can be split into two half-reactions:$1) \ Cu^{+} 	o Cu^{2+} + e^{-} \ (E^o_{o

Vedclass · Accepted Answer

$0.38$

Vedclass · Answer

(D) The disproportionation reaction is $2Cu^{+} 	o Cu^{2+} + Cu$.This can be split into two half-reactions:$1) \ Cu^{+} 	o Cu^{2+} + e^{-} \ (E^o_{ox} = -E^o_{Cu^{2+}/Cu^{+}} = -0.15 \ V)$$2) \ Cu^{+} + e^{-} 	o Cu \ (E^o_{red} = E^o_{Cu^{+}/Cu})$First,we calculate $E^o_{Cu^{+}/Cu}$ using the relation $\Delta G^o_{total} = \Delta G^o_1 + \Delta G^o_2$:$1 	imes F 	imes E^o_{Cu^{+}/Cu} = 2 	imes F 	imes E^o_{Cu^{2+}/Cu} - 1 	imes F 	imes E^o_{Cu^{2+}/Cu^{+}}$$E^o_{Cu^{+}/Cu} = 2(0.34) - 0.15 = 0.68 - 0.15 = 0.53 \ V$.Now,for the disproportionation reaction:$E^o_{cell} = E^o_{red} + E^o_{ox} = E^o_{Cu^{+}/Cu} - E^o_{Cu^{2+}/Cu^{+}}$$E^o_{cell} = 0.53 \ V - 0.15 \ V = 0.38 \ V$.

$Cu^{+}$ ion is not stable in aqueous solution because of disproportionation reaction. $E^o$ value for disproportionation of $Cu^{+}$ is .............. $V$ (Given $E^o_{Cu^{2+}/Cu^{+}} = 0.15 \ V$,$E^o_{Cu^{2+}/Cu} = 0.34 \ V$)

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