Question

A button cell used in watches functions as following
Zn(s) + Ag₂O(s) + H₂O(l) ⇌ 2Ag(s) + Zn²⁺ (aq) + 20H^- (aq)
If half cell potentials are
Zn²⁺(aq) + 2e^- → Zn(s); E° = -0.76 V
Ag₂O(s) + H₂O(l) + 2e^- → 2Ag(s) + 20H^-(aq),
E° = 0.34 V
The cell potential will be

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

Answer 1

The Correct Option is A

To find the cell potential of the given electrochemical cell, we use the formula for calculating the standard cell potential:

E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}

In this scenario, the half-reactions provided are:

• Anode reaction: \text{Zn}^{2+}(aq) + 2e^{-} \rightarrow \text{Zn}(s) with E^{\circ} = -0.76 \, \text{V}
• Cathode reaction: \text{Ag}_{2}\text{O}(s) + \text{H}_{2}\text{O}(l) + 2e^{-} \rightarrow 2\text{Ag}(s) + 2\text{OH}^{-}(aq) with E^{\circ} = 0.34 \, \text{V}

From these equations, Zn/Zn²⁺ serves as the anode, and Ag₂O/Ag as the cathode in a galvanic cell.

Plug the given values into the standard cell potential formula:

E_{\text{cell}}^{\circ} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V}

Thus, the standard cell potential for this electrochemical process is 1.10 V.

The correct answer is 1.10 V, which matches with the given correct option.

Explanation:

• In this electrochemical cell, the metal zinc is oxidized (loses electrons), while silver oxide (Ag₂O) is reduced (gains electrons).
• The positive potential indicates a spontaneous reaction, which is typical for galvanic cells.

Concepts Used:

• The standard electrode potential measures the individual potential of a reversible electrode at standard conditions, which is crucial in predicting the direction of electron flow.
• The overall cell potential is the difference between the reduction potential of the cathode and that of the anode.