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Calculate the electromotive force (→ potential) under the standard conditions of this battery:
$E^o_{battery}$ $ =$ $ 0.81-(0.75)=0.06\; V$
$n\; =\; 1$ indeed $Ag^{+}$ + 1$e^-$ + $Fe^{2+}$ - 1$e^- $ $\leftrightarrows$ $Fe^{3+}+Ag$
$E$ $ =$ $ E^o_{battery}-\frac{0.059}{1}log\frac{[Fe^{3+}]}{[Fe^{2+}][Ag^{+}]}$ $ = $ $0.06-\frac{0.059}{1}log\frac{0.20}{0.10\cdot 1.0} $ $=$ $0.042\;V$