Question:
Consider the equilibrium obtained by electrically connecting zinc-amalgam (Zn(Hg)) and HgO electrodes in mercury cell, $ \text{Zn(Hg)} + \text{HgO(s)} \rightleftharpoons \text{ZnO(s)} + \text{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?}
Consider the equilibrium obtained by electrically connecting zinc-amalgam (Zn(Hg)) and HgO electrodes in mercury cell, $ \text{Zn(Hg)} + \text{HgO(s)} \rightleftharpoons \text{ZnO(s)} + \text{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?}
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At equilibrium of electrochemical cell, potential difference becomes zero.
Updated On: May 1, 2026
- Zn(Hg) electrode potential is equal to HgO electrode potential
- Zn(Hg) electrode potential is more than HgO electrode potential
- HgO electrode potential is more than Zn(Hg) electrode
- Cell voltage at above said equilibrium is 1.35 V
- Both (C) and (D)
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The Correct Option is A
Solution and Explanation
Concept:
At equilibrium: \[ E_{cell} = 0 \] Also: \[ E_{cell} = E_{cathode} - E_{anode} \]
Step 1: Apply equilibrium condition.
\[ E_{cell} = 0 \Rightarrow E_{cathode} = E_{anode} \]
Step 2: Interpretation.
Thus both electrode potentials are equal.
Step 3: Conclusion.
\[ E_{\text{Zn(Hg)}} = E_{\text{HgO}} \]
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