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

Question

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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