Electrolytic and galvanic cells, electrode potentials
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See all →- Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is:(Given : Molar mass of Cu : 63 g mol^-1, 1F = 96487 C) Answer: 3. 0.315 g Key idea: Q=It , then use Faraday's law for mathrmCu²⁺+2e^-→ Cu . Q=9.6487×100=964.87 C, so moles of e^- = 964.87/96487 =... 1
- Match List I with List II.List I (Conversion): A. 1 mol of H_2O to O_2 B. 1 mol of MnO_4^- to Mn^2+ C. 1.5 mol of Ca from molten CaCl_2 D. 1 mol of FeO to Fe_2O_3List II (Number of Faraday required): I. 3F II. 2F III. 1F IV. 5FChoose the correct answ… Answer: 2. A-II, B-IV, C-I, D-III Key idea: Faradays = moles of electrons transferred. A: 2H₂O→ O₂+4H^++4e^- , so 1 mol H₂O needs 2 F (II).... 1
- At 298 K, the standard electrode potentials of Cu^2+/Cu, Zn^2+/Zn, Fe^2+/Fe and Ag^+/Ag are 0.34 V, -0.76 V, -0.44 V and 0.80 V, respectively.On the basis of standard electrode potential, predict which of the following reaction can not occur ? Answer: 1. 2CuSO₄(aq) + 2Ag(s) → 2Cu(s) + Ag₂SO₄(aq) Key idea: a metal can displace another metal from its salt only if its standard reducti... 1
- The number of Faradays(F) required to produce 20 g of calcium from molten CaCl_2 (Atomic mass of Ca = 40 g mol^-1) is : Answer: 1. 1 Electrode reaction: mathrmCa²⁺+2e^-→ Ca . One mole of Ca needs 2 F. Moles of Ca =20/40=0.5 mol. Charge =0.5×2=1 F. Answer: opti... 1
- On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be : Answer: 2. Oxygen gas In dilute H₂SO₄ with inert Pt electrodes, water is oxidised at the anode more easily than the sulphate ion. Anode: 2H₂... 1
- During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is Answer: 1. 110 minutes Key idea: 2Cl^-→ Cl₂+2e^- , so 1 mol Cl₂ needs 2 F. Q=0.10×2×96500=19300\ C t=Q/I=19300/3≈6433\ s≈107\ min≈110\ min T... 1