A galvanic cell transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. In a galvanic cell, electrons flow from

For $n$-electron redox reactions of the type $\mathrm{aA}+\mathrm{bB}⟶\mathrm{cC}+\mathrm{dD},$ the cell potential can be expressed as

What will be the $\mathrm{E}_{\text {cell }}$ for the given cell: $\mathrm{Zn}/{\mathrm{Zn}}^{2+}(0.1\mathrm{M})//{\mathrm{Cu}}^{2+}(0.01\mathrm{M})/\mathrm{Cu}$ 

Given: ${{\mathrm{E}}^{\mathrm{o}}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}=-0.76 \mathrm{V}$ and ${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\mathrm{o}}=0.34 \mathrm{V}$ 

Also predict whether the reaction is spontaneous or nonspontaneous.

According to the expression $\mathrm{\Delta G}°=-\mathrm{nFE}°$, the cell reaction is spontaneous when

(Notations and symbols carry their usual meanings)