In recharging a lead-acid accumulator, the reaction at the cathode can be represented as
Pb2+(aq) + SO2-4(aq) →PbSO4(s)
Pb2+(aq) + 2e- →Pb(s)
Pb2+(aq) + 2H2O(l) → PbO2(s) + 4H+(aq) + 4e-
Pb(s) → Pb2+(aq) + 2e-
Explanation
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Ok Lemme clear all doubts
The lead-acid accumulator is commonly used as a car battery. It is a secondary or storage cell as it must be charged by passing a direct current through it. The charged cell can then produce an electric current when required. The cathode of a fully charged accumulator is lead(IV) oxide, PbO₂, and the anode is metallic lead. The electrolyte is dilute tetraoxosulphate(VI) acid solution, H₂SO₄
DISCHARGING THE CELL
At The Anode: During the process of discharging, the lead atoms release two electrons to become oxidised to lead(II)ions Pb²⁺, which then combine with the tetraoxosulphate(VI) ions SO₄²⁻, in the electrolyte to become deposited on the anode as lead(II) tetraoxosulphate(VI), PbSO₄
Pb --> Pb²⁺ + 2e⁻
Pb²⁺ + SO₄²⁻ --> PbSO₄
At The Cathode: The electrons from the anode are accepted at the cathode where the lead(IV) oxide and the hydrogen ions from the electrolyte undergo reduction to produce lead(II) ions and water
PbO₂ + 4H⁺ + 2e⁻ --> Pb²⁺ + 2H₂O
The lead(II) ions then combine with the tetraoxosulphate(VI) ions from the electrolyte to become deposited at the cathode as lead(ii) tetraoxosulphate(VI)
Pb²⁺ + SO₄²⁻ --> PbSO₄(s)
When both electrodes are completely covered with lead(ii) tetraoxosulphate(VI) deposits, the lead-acid accumulator will stop discharging a current. To make it produce a current again, it will have to be recharged.
RECHARGING THE CELL
During the process of recharging, the redox reactions at the respective electrodes in the cell are reversed. The electrode which is the anode in the discharged cell becomes the cathode, while the cathode becomes the anode.
At the Cathode
PbSO₄ --> Pb²⁺ + 2e⁻ --> Pb(s)
NB: SO₄²⁻ is also produced
At the Anode
PbSO₄ --> Pb²⁺ --> PbO₂(s) + 4H⁺ + 2e-
NB: SO₄²⁻ is also produced
The question asked what the reaction at the CATHODE is when RECHARGING the cell. So the answer is B
The answer is B ohhhhh go read your textbook, it is not all of myschool's answers that are correct
𝐃𝐈𝐒𝐂𝐇𝐀𝐑𝐆𝐈𝐍𝐆
𝐀𝐧𝐨𝐝𝐞: 𝐏𝐛 -> 𝐏𝐛²⁺ + 𝟮𝐞⁻ → 𝐏𝐛𝐒𝐎₄
𝐂𝐚𝐭𝐡𝐨𝐝𝐞: 𝐏𝐛𝐎₂ + 𝟰𝐇⁺ + 𝟮𝐞⁻ → 𝐏𝐛²⁺ + 𝟮𝐇₂𝐎 → 𝐏𝐛𝐒𝐎₄
𝐑𝐄𝐂𝐇𝐀𝐑𝐆𝐈𝐍𝐆
𝐂𝐚𝐭𝐡𝐨𝐝𝐞: 𝐏𝐛𝐒𝐎₄ → 𝐏𝐛²⁺ + 𝐒𝐎₄²⁻ → 𝐏𝐛²⁺ + 𝟮𝐞⁻ → 𝐏𝐛
𝐀𝐧𝐨𝐝𝐞: 𝐏𝐛𝐒𝐎₄ → 𝐏𝐛²⁺ + 𝐒𝐎₄²⁻ → 𝐏𝐛²⁺ + 𝟮𝐇₂𝐎 → 𝐏𝐛𝐎₂ + 𝟰𝐇⁺ + 𝟮𝐞⁻
𝐁
Hello guys, listen up.
Since we are asked about RECHARGING and we all know when the accumulator is fully DISCHARGED, both electrodes are now covered with PbSO4, which is Pb2+ and sulphate ion, therefore when RECHARGING, the Pb2+ at the electrodes convert to PbO2 (at the anode) or Pb (at the cathode), therefore restoring the main nature of the rods back..
So all I'm tryna say is that, during RECHARGING, lead ions are converted to metallic lead or lead perixide.. from the options given, the answer is either B or C, right?? So, since originally the anode is PbO2 and cathode is Pb, therefore B is the answer.
This should help..
After many confusion and I myself confused 100%,I made a tough research on this, D is right. Reduction takes place at the cathode while oxidation takes place at d anode.
option D is for discharging d cell at d anode. B shud be d ryt answer.. check Ababio chemistry pg 217
Check Understanding chemistry for this and all will be revealed, seriously is no one seeing that the answer is D
