Question

Consider the equilibrium obtained by electrically connecting zinc-amalgam $Zn(Hg)$ and $HgO$ electrodes in mercury cell,
$Au(Hg)+HgO(s)\rightleftharpoons ZnO(s)+Hg(l)$
Under this equilibrium, what is the relation between the potential of the $Zn(Hg)$ and $HgO$ electrodes measured against the standard hydrogen electrode?

• A. Zn(Hg) electrode potential is equal to HgO electrode potential
• B. Zn(Hg) electrode potential is more than HgO electrode potential
• C. HgO electrode potential is more than Zn(Hg) electrode
• D. Cell voltage at above said equilibrium is 1.35 V
• E. Both (C) and (D)

Answer 1

Answer: (A)

Since, at equilibrium $E_{(\text{cell })}^{\circ }=$ zero

and $E_{\text{(cell) }}^{\circ }=E_{(\text{cathode })}^{\circ }-E_{(\text{anode })}^{\circ }$

at equilibrium

$E_{(\text{cathode })}^{\circ }=E_{(\text{anode })}^{\circ }\text{ i.e }$

electrode potential of $Zn(Hg)$ electrode must be equal to the $HgO$ electrode.