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A solution of \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 is electrdyzed for 'X' min a current of 2.8 A to deposit 0.3485 g of \mathrm{Fe}. The vaule of X is:-

Option: 1

20 min


Option: 2

15 min


Option: 3

10 min


Option: 4

25 min


Answers (1)

best_answer

\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \rightarrow 2 \mathrm{Fe}

\mathrm{W = z ~i ~t \\ }

\mathrm{W=\left(\frac{E}{96500}\right) i ~t }

\mathrm{0.3485=\left(\frac{56}{3 \times 96500}\right) \times 2.8 \times t }

\mathrm{t=643.43 \mathrm{~sec}}

\mathrm{=10 \mathrm{~min}}

Posted by

Devendra Khairwa

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