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Current Electricity question

2004 · Shift 0 · Q137
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Current Electricity question

2004 · Shift 0 · Q137

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The electrochemical equivalent of a metal is 3.35109−7kg{3.35109^{ - 7}}kg3.35109−7kg per Coulomb. The mass of the metal liberated at the cathode when a 3A3A3A current is passed for 222 seconds will be
  1. A
    6.6×1057/kg6.6 \times {10^{57}}/kg6.6×1057/kg
  2. B
    9.9×10−7 kg9.9 \times {10^{ - 7}}\,kg9.9×10−7kg
  3. C
    19.8×10−7 kg19.8 \times {10^{ - 7}}\,kg19.8×10−7kg
  4. D
    1.1×10−7 kg1.1 \times {10^{ - 7}}\,kg1.1×10−7kg
View written solutionFree

Correct answer: C

  1. Use the relation for electrochemical deposition

    The mass deposited is given by m=ZQm = ZQm=ZQ where:

    • Z=3.35109×10−7 kg/CZ = 3.35109 \times 10^{-7}\,\text{kg/C}Z=3.35109×10−7kg/C is the electrochemical equivalent,
    • Q=ItQ = ItQ=It is the charge passed.

  2. Calculate the charge passed

    Given: I=3 A,t=2 sI = 3\,\text{A}, \quad t = 2\,\text{s}I=3A,t=2s

    So, Q=It=3×2=6 CQ = It = 3 \times 2 = 6\,\text{C}Q=It=3×2=6C

  3. Calculate the mass liberated

    m=ZQ=(3.35109×10−7)(6)m = ZQ = (3.35109 \times 10^{-7})(6)m=ZQ=(3.35109×10−7)(6)

    m=20.10654×10−7 kgm = 20.10654 \times 10^{-7}\,\text{kg}m=20.10654×10−7kg

    m=2.010654×10−6 kgm = 2.010654 \times 10^{-6}\,\text{kg}m=2.010654×10−6kg

  4. Match with the options

    20.10654×10−7 kg≈19.8×10−7 kg20.10654 \times 10^{-7}\,\text{kg} \approx 19.8 \times 10^{-7}\,\text{kg}20.10654×10−7kg≈19.8×10−7kg

    So the closest option is: Option C

  5. Comparison with stored answer

    Stored correct answer: C

    My derived answer: C

    Hence, they agree.

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