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Electrochemistry question

2019 · 9 Apr · Shift 2 · Q8
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Electrochemistry question

2019 · 9 Apr · Shift 2 · Q8

JEE MainChemistryElectrochemistryMCQ+4 / −1
A solution of Ni(NO3)2Ni(NO_3)_2Ni(NO3​)2​ is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of NiNiNi will be deposited at the cathode?
  1. A
    0.10
  2. B
    0.15
  3. C
    0.20
  4. D
    0.05
View written solutionFree

Correct answer: D

  1. Identify the cathode reaction

For electrolysis of Ni(NO3)2Ni(NO_3)_2Ni(NO3​)2​ using platinum electrodes, the cation reduced at the cathode is:

Ni2++2e−→Ni(s)Ni^{2+} + 2e^- \rightarrow Ni(s)Ni2++2e−→Ni(s)

So, 2 moles of electrons are required to deposit 1 mole of Ni.

  1. Use Faraday concept

Given electricity passed =0.1= 0.1=0.1 Faraday.

Since:

1 Faraday=1 mole of electrons1 \text{ Faraday} = 1 \text{ mole of electrons}1 Faraday=1 mole of electrons

therefore,

0.1 Faraday=0.1 mole of electrons0.1 \text{ Faraday} = 0.1 \text{ mole of electrons}0.1 Faraday=0.1 mole of electrons

  1. Calculate moles of nickel deposited

From the electrode reaction:

2 mol e− deposit 1 mol Ni2 \text{ mol } e^- \text{ deposit } 1 \text{ mol Ni}2 mol e− deposit 1 mol Ni

So,

0.1 mol e− will deposit 0.12=0.05 mol Ni0.1 \text{ mol } e^- \text{ will deposit } \frac{0.1}{2} = 0.05 \text{ mol Ni}0.1 mol e− will deposit 20.1​=0.05 mol Ni

  1. Match with the options

0.05 mol0.05 \text{ mol}0.05 mol

So the correct option is:

D: 0.05

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