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Electrochemistry question

2014 · Q93
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Electrochemistry question

2014 · Q93

NEETChemistryElectrochemistryMCQ+4 / −1
When 0.1 mol MnO42−_4^{2 - }42−​ is oxidised the quantity of electricity required to completely oxidise MnO42−_4^{2 - }42−​ to MnO4−_4^ -4−​ is
  1. A
    96500 C
  2. B
    2 ×\times× 96500 C
  3. C
    9650 C
  4. D
    96.50 C
View written solutionFree

Correct answer: C

MnO42−0.1 mole−6\mathop {MnO_4^{2 - }}\limits_{0.1\,mole}^{ - 6} 0.1moleMnO42−​−6​ →\to→ MnO4−−7\mathop {MnO_4^ - }\limits^{ - 7} MnO4−​−7​ + e-

Quantity of electricity required = 0.1F

= 0.1 × 96500 = 9650 C

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