Question

A button cell used in watches function as following Zn(s) + Ag₂O(s) + H₂O(l) \(\rightleftharpoons\) 2Ag(s) + Zn²⁺(aq) + 2OH^-(aq). If half cell potentials are \({Zn^{2+}_{ (aq)} + 2e\to Zn(s); E^{\circ} = -0.76V}\) \({Ag2O_{(s)} + H2O_{(l)} + 2e\to 2Ag_{(s)} + 2OH^{-}_{(aq)};}\) \(E^{\circ} = 0.34V\)

• A. 1.34 V
• B. 1.10 V
• C. 0.42 V
• D. 0.84 V

Answer 1

The Correct Option is B

To determine the overall cell potential (E⁰), we use the given half-cell reactions and their standard electrode potentials. In electrochemistry, the cell potential is calculated using the formula:

E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}

Here, the cathode is the reduction reaction and the anode is the oxidation reaction. For the given cell, we have:

1. Reduction half-reaction at the cathode:
   • {Ag_2O_{(s)} + H_2O_{(l)} + 2e^- \to 2Ag_{(s)} + 2OH^-_{(aq)}}; \, E^{\circ} = 0.34 \, V
2. Oxidation half-reaction at the anode:
   • {Zn_{(s)} \to Zn^{2+}_{(aq)} + 2e^-}; \, E^{\circ} = -0.76 \, V

Plug the values into the cell potential equation:

E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 \, V - (-0.76 \, V)

Calculating this gives:

E^{\circ}_{cell} = 0.34 \, V + 0.76 \, V = 1.10 \, V

Therefore, the cell potential is 1.10 V.

Conclusion: The correct answer is 1.10 V. This is reached by understanding the cell reactions, determining which is reduced and which is oxidized, and then applying the standard cell potential formula.