In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq). Determine ΔrG° and E° for the reaction.
E° = 1.10 V and ΔrG° = -212.27 kJ mol⁻¹
Explanation
The question requires determining standard cell potential and Gibbs free energy for the button cell reaction. The half-reactions are: Oxidation at anode: Zn(s) → Zn²⁺(aq) + 2e⁻ with E°(Zn²⁺/Zn) = -0.76 V; Reduction at cathode: Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq) with E°(Ag₂O/Ag) = +0.34 V. The standard cell potential is calculated as E°(cell) = E°(cathode) - E°(anode). Using the formula ΔrG° = -nFE°(cell) from the context (equation 2.16), where n = 2 and F = 96487 C mol⁻¹, we can calculate ΔrG°.
Solution Steps
Step 1: Identify half-reactions Oxidation (anode): Zn(s) → Zn²⁺(aq) + 2e⁻ Reduction (cathode): Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq)
Step 2: Determine standard electrode potentials E°(Zn²⁺/Zn) = -0.76 V E°(Ag₂O/Ag) = +0.34 V
Step 3: Calculate E°(cell) E°(cell) = E°(cathode) - E°(anode) E°(cell) = 0.34 - (-0.76) = 1.10 V
Step 4: Calculate ΔrG° Using ΔrG° = -nFE°(cell) where n = 2 and F = 96487 C mol⁻¹ ΔrG° = - C mol⁻¹ × 1.10 V ΔrG° = -212271.4 J mol⁻¹ -212.27 kJ mol⁻¹