What is the number of moles of electrons gained by n mole PbO _ 2(s) when it is converted to PbSO _ 4(s) at discharging of lead accumulator?

Step-by-step solution

During the discharging of a lead accumulator, the reduction half-reaction at the cathode is given by: PbO _ 2(s) + SO ₄²⁻_ (aq) + 4 H ⁺_ (aq) + 2e⁻ PbSO _ 4(s) + 2 H ₂ O _ (l) The oxidation state of Pb in PbO ₂ is +4 . The oxidation state of Pb in PbSO ₄ is +2 . The change in oxidation state of Pb is from +4 to +2 , which corresponds to a gain of 2 electrons per molecule of PbO ₂ . Thus, 1 mole of PbO ₂ gains 2 moles of electrons. Therefore, n moles of PbO ₂ will gain 2n moles of electrons. Answer: 2n