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We consider the following redox system: $Ag^{+}$ // $Ag$ $E^o$= $0.81 \;V$ What is his → potential at $90^oC$ for $[Ag^{+}]$ $=$ $0.1 \frac{mol}{L}$ ?
$E$ $ =$ $ E^o+\frac{RT}{nF}ln[Ag^{+}]$
$n\; =\; 1$ because $Ag^{+}$ + 1 $e^- $ $\leftrightarrows$ $Ag$
$\frac{RT}{nF}\;$$ =$$\;\frac{8.3145\cdot (273.15+90)}{1\cdot96 485}$$ = $$0.0313 $
Taking into account the conversion ln → log: $E$ $ =$ $ 0.75+\frac{2.3\;0.0313}{1}log(0.1)$ $ =$ $0.68\;V$