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  •  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 oxid

 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)

Option: 1

1+


Option: 2

2+


Option: 3

3+


Option: 4

4+


Answers (1)

best_answer

i = 10 A, t = 2hr

\text { No. of moles }=\frac{I t}{96500 \times(n-\text { factor })}

\\\therefore\mathrm{\textrm{ moles of e}^- = \frac{10\times 2\times 60\times 60}{96500\times n-factor}}\\\\\therefore\mathrm{\textrm{0.25}= \frac{10\times 2\times 60\times 60}{96500\times X}} \\\\\therefore 0.75 = 0.25 \times(X) \\\\\Rightarrow X=3 \\\\\therefore\mathrm{\textrm{Metal X is present in the form of X}^{3+}}

Therefore, the correct option is (3).

Posted by

Pankaj Sanodiya

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