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Calculate the electromotive force (→ potential) under the standard conditions of a battery containing the electrodes $Zn^{2+}$ // $Zn$ $E^o$= $-0.76 \;V$ $Cu^{2+}$ // $Cu$ $E^o$= $0.34 \;V$ knowing that $[Cu^{2+}]=10^{-9} \frac{mol}{L}$ $[Zn^{2+}]=0.1 \frac{mol}{L}$
$E^o_{battery}$ $ =$ $ 0.34-(-0.76)=1.10\; V$
$n\; =\; 2$ indeed $Cu^{2+}$ + 2$e^-$ + $Zn$ - 2$e^- $ $\leftrightarrows$ $Zn^{2+}+Cu$
$E$ $ =$ $ E^o_{battery}-\frac{0.059}{2}log\frac{[Zn^{2+}]}{[Cu^{2+}]}$ $ = $ $1.10-\frac{0.059}{2}log\frac{0.1}{10^{-9}} $ $=$ $0.86\;V$