For the reduction of silver ions with copper | vedclass

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(A) The standard Gibbs energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$The cell reaction is: $2 Ag^{+} + Cu \rightarrow

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(A) The standard Gibbs energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$The cell reaction is: $2 Ag^{+} + Cu ightarrow Cu^{2+} + 2 Ag$Here,the number of electrons transferred,$n = 2$.Given,$E^{\circ}_{cell} = + 0.46 \ V$ and $F = 96500 \ C \ mol^{-1}$.Substituting the values: $\Delta G^{\circ} = -2 	imes 96500 	imes 0.46 \ J \ mol^{-1}$$\Delta G^{\circ} = -88780 \ J \ mol^{-1}$Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -88.78 \ kJ \ mol^{-1}$Rounding to the nearest value,we get $\Delta G^{\circ} \approx -89.0 \ kJ \ mol^{-1}$.

For the reduction of silver ions with copper metal,the standard cell potential was found to be $+ 0.46 \ V$ at $25^oC.$ The value of standard Gibb's energy,$\Delta G^o$ will be .......... $kJ.$
$(F = 96500 \ C \ mol^{-1})$

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For the reduction of silver ions with copper metal,the standard cell potential was found to be $+ 0.46 \ V$ at $25^oC.$ The value of standard Gibb's energy,$\Delta G^o$ will be .......... $kJ.$$(F = 96500 \ C \ mol^{-1})$