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Standard electrode potential for the cell with cell reaction $Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)$ is 1.1 V. Calculate the standard Gibbs energy change for the cell reaction. (Given $F=96487$ C mol$^{-1}$)
Detailed Solution
$n=2$. $\Delta G^\circ=-nFE^\circ_{cell}=-2\times96487\times1.1=-212271$ J mol$^{-1}\approx-212.27$ kJ mol$^{-1}$.