The zinc–silver(I) oxide button cell is used in hearing aids, the overall cell reaction being represented by the equation: Ag2O(s) + Zn(s) + H2O(l) → 2Ag(s) + Zn(OH)2(s). If the cell is to deliver a continuous current of 0.10 mA for a period of 10 weeks (1680 hours), the mass of silver(I) oxide required in the cell is

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The zinc–silver(I) oxide button cell is used in hearing aids, the overall cell reaction being represented by the equation: Ag2O(s) + Zn(s) + H2O(l) → 2Ag(s) + Zn(OH)2(s). If the cell is to deliver a continuous current of 0.10 mA for a period of 10 weeks (1680 hours), the mass of silver(I) oxide required in the cell is

BlackTom AI · Accepted Answer

To start, translate the given data into electrical quantities: the current is 0.10 mA, which is 0.00010 A, and the duration is 1680 hours. Converting time to seconds gives 1680 h × 3600 s/h = 6,048,000 s. The total electric charge delivered is Q = I × t = 0.00010 A × 6,048,000 s = 604.8 C.

Next, relate the charge to moles of electrons with Faraday's constant: n(e−) = Q / F = 604.8 C / 96485 C/mol ≈ 0.00627 mol e−.

For the cell reaction, Ag2O is reduced to Ag by gaining electrons. The half-reaction can be written as Ag2O + H2O + 2 e− → 2 Ag + 2 OH−, meaning 2 moles of electrons are needed to reduce 1 mole of Ag2O to 2 moles of Ag. Therefore, the moles of Ag2O that were reduced (and thus required) are n(Ag2O) = n(e−) / 2 ≈ 0.00627 / 2 ≈ 0.00314 mol.

Now calculate the mass of Ag2O required: molar mass of Ag2O = 2 × 107.87 g/mol (Ag) + 16.00 g/mol (O) = 231.74 g/mol. Mass = n × M = 0.00314 mol × 231.74 g/mol ≈ 0.728 g, which rounds to 0.73 g.

Why the other options are not correct:
- 1.5 g would imply a much larger amount of Ag2O being reduced than the given current over the stated time; the calculated charge corresponds to far less material.
- 0.20 g would correspond to a smaller amount of Ag2O than required by the total electrons transferred; using the same stoichiometry shows this underestimates the mass.
- 0.40 g is also too small; it would require fewer moles of Ag2O than the electron budget allows, so it would not supply enough silver for the given current and time.
- The calculation above demonstrates that 0.73 g is the mass consistent with Q = 604.8 C and the 2 e− per Ag2O reduction step; any other value misaligns with the electron-to-mass stoichiometry of the cell reaction.

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