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Calculate the electromotive force (→ potential) under the standard conditions of this battery: Pt|$Cl_2$ // $Cl{-}$ $E^o$= $1.36 \;V$ $Ni^{2+}$ // $Ni$ $E^o$= $-0.25 \;V$ knowing that $P_{Cl_2}=1 atm$ $[Cl^{-}]=0.2 \frac{mol}{L}$ $[Ni^{2+}]=0.01 \frac{mol}{L}$
$E^o_{battery}$ $ =$ $ 1.36-(0.25)=1.61\; V$
$n\; =\; 2$ indeed $Cl_2$ + 2$e^-$ + $Ni$ - 2$e^- $ $\leftrightarrows$ $Ni^{2+}+2Cl^{-}$
$E$ $ =$ $ E^o_{battery}-\frac{0.059}{2}log\frac{[Ni^{2+}][Cl^{-}]^2}{ P_{Cl_2}}$ $ = $ $1.61-\frac{0.059}{2}log\frac{0.01\cdot 0.2^2}{1} $ $=$ $1.71\;V$