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.