If the cell reaction is spontaneous 

The electrical work of the electrochemical cell is given as 

$∆G_{ }^o=-nF{E}_{cell}^o$

$$\begin{gathered} \mathrm{if} \\ ∆{\mathrm{G}}^{\mathrm{o}}= 0 (\mathrm{the} \mathrm{cell} \mathrm{is} \mathrm{at} \mathrm{equiliobrium})\hspace{0.17em} \\ ∆{\mathrm{G}}^{\mathrm{o}}= \mathrm{negative} (\mathrm{the} \mathrm{cell} \mathrm{is} \mathrm{spentaneous})\hspace{0.17em} \\ ∆{\mathrm{G}}^{\mathrm{o}}= \mathrm{positive} (\hspace{0.17em}\mathrm{the} \mathrm{cell} \mathrm{is} \mathrm{non}-\mathrm{spentaneous})\hspace{0.17em} \end{gathered}$$

If ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}= \mathrm{is} \mathrm{positive} , \mathrm{then} ∆{\mathrm{G}}_{ }^{\mathrm{o}} \mathrm{will} \mathrm{be} \mathrm{negative} , \mathrm{then} \mathrm{the} \mathrm{electrochemical} \mathrm{cell} \mathrm{will} \mathrm{be} \mathrm{spontaneous}.$