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Calculate the electromotive force (→ potential) under the standard conditions of a battery containing the electrodes $Fe^{2+}$ // $Fe$ $E^o$= $-0.44 \;V$ $Cd^{2+}$ // $Cd$ $E^o$= $-0.40 \;V$ knowing that $[Cd^{2+}]=10^{-3} \frac{mol}{L}$ $[Fe^{2+}]=1 \frac{mol}{L}$
$E^o_{battery}$ $ =$ $ -0.40-(-0.44)=0.04\; V$
$n\; =\; 2$ indeed $Cd^{2+}$ + 2$e^-$ + $Fe$ - 2$e^- $ $\leftrightarrows$ $Fe^{2+}+Cd$
$E$ $ =$ $ E^o_{battery}-\frac{0.059}{2}log\frac{[Fe^{2+}]}{[Cd^{2+}]}$ $ = $ $0.04-\frac{0.059}{2}log\frac{1}{10^{-3}} $ $=$ $-0.049\;V$
Where is the + pole?
$ Fe^{2+} $ // $ Fe $ is the + pole, although its standard potential is smaller!