Q2 / 90 · JEE Main 2014 (09 Apr Online) · Chemistry

A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is:

F=96,500 C

  • A
    1+
  • B
    4+
  • Correct answer
    C
    3+
  • D
    2+

Solution

Quantity of Electricity  = 1 0 × 3 6 0 0 × 2 9 6 5 0 0
                                   =0.746≈0.75F.
If oxidation state is n+ for 1 mol deposition of a metal:
Electricity=nF (Faradays)
For 0.250 mol=0.250 nF
∴ 0.250 nF=0.75 F
n=3

Hence, the correct answer is option: 'C'.

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