In a dry cell,which element acts as the negative electrode?

For the reaction $2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightarrow 2 Fe^{2+}_{(aq)} + I_{2(s)}$,the magnitude of the standard molar free energy change,$\Delta_{r} G_{m}^{\circ} = -$ ........... $kJ$ (Round off to the Nearest Integer). $\left[ E_{Fe^{2+} / Fe_{(s)}}^{\circ} = -0.440 \ V; \ E_{Fe^{3+} / Fe_{(s)}}^{\circ} = -0.036 \ V; \ E_{I_{2} / 2I^{-}}^{\circ} = 0.539 \ V; \ F = 96500 \ C \ mol^{-1} \right]$

Given the standard electrode potentials $E^{0}_{Fe^{2+}|Fe} = -0.44 \, V$ and $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.77 \, V$. If $Fe^{2+}$,$Fe^{3+}$,and $Fe$ are all present together,then:

If $\Lambda _{NaCl}^{o}=126\,S\,cm^{2}mol^{-1}$,$\Lambda _{KBr}^{o}=152\,S\,cm^{2}mol^{-1}$,and $\Lambda _{KCl}^{o}=150\,S\,cm^{2}mol^{-1}$,then find $\Lambda ^{o}_{NaBr}$ in $S\,cm^{2}mol^{-1}$.

Give the anodic,cathodic,and overall reactions of a hydrogen-oxygen fuel cell.

The reference electrode is made by using