Question of 115
Q.(a) Write the equations for the reactions occurring at the cathode and anode during the charging of a lead storage cell (lead accumulator).
(b) Using a platinum electrode, a current of 5 amperes was passed through a solution of Ni(NO3)2 for 20 minutes. What quantity of nickel will be deposited at the cathode?
(c) What is a cell constant?
OR
(a) What is the relationship between the specific conductivity and molar conductivity of a solution?
(b) When a current of 0.6 amperes is passed through a copper sulphate (CuSO4) solution for 40 minutes, 0.2964 g of copper is deposited. What is the electrochemical equivalent of copper?
(c) Write the Nernst equation for the following cell: Mg(s) | Mg2+(0.001 M) || Cu2+(0.0001 M) | Cu(s)
(d) Write the cell notation/representation for the cell corresponding to the following reaction: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Tripura TbseHigher Secondary (+2 Stage) Examination 2026Subjective· 5mImportance★★★★★
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Start your 14-day free trial to unlock the full solution →Charging a lead-acid battery reverses the discharge reactions, regenerating Pb at the cathode and PbO2 at the anode from PbSO4; using Faraday's law, about 1.83 g of nickel is deposited by the given current and time; the cell constant relates measured conductance to specific conductance. (Answering the primary version of this OR question, per the solving contract.)
- During charging of a lead storage cell (lead-acid accumulator), an external current is forced through the cell in the reverse direction to discharge, converting lead sulphate (PbSO4) deposited on both electrodes back into the active materials: At the cathode (reduction): PbSO4(s) + 2e- -> Pb(s) + SO4^2-(aq) At the anode (oxidation): PbSO4(s) + 2H2O(l) -> PbO2(s) + 4H+(aq) + SO4^2-(aq) + 2e- Overall (charging): 2PbSO4(s) + 2H2O(l) -> Pb(s) + PbO2(s) + 2H2SO4(aq)
- Current I = 5 A, time t = 20 min = 20 x 60 = 1200 s Charge passed, Q = I x t = 5 x 1200 = 6000 C Ni2+ + 2e- -> Ni(s); 2 moles of electrons (2F = 2 x 96500 = 193000 C) deposit 1 mole of Ni (molar mass approximately 58.7 g/mol) Moles of electrons passed = Q/F = 6000/96500 = 0.06218 mol Moles of Ni deposited = 0.06218/2 = 0.03109 mol Mass of Ni deposited = 0.03109 x 58.7 = approximately 1.825 g …
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