CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

Question : 12

Total: 26

(a) The cell in which the following reaction occurs:
2Fe3+(aq)+2I−(aq)→2Fe2+(aq) +I2( s)
has E°‌cell ‌=0.236V at 298K. Calculate the standard Gibb's energy of the cell reaction.
(Given : 1F=96,500Cmol−1 )
(b) How many electrons flow through a metallic wire if a current of 0.5A is passed for 2 hours ?
(Given : 1F=96,500Cmol−1 )

Solution:

‌ (a) ‌ ‌2Fe3++2e−→2Fe2+
2I−→I2+2e−
For the given cell reaction, n=2.
∆G∘=−nFE°‌cell ‌
‌=−2×96500×0.236
‌=−45548Jmol−1
‌=−45.55‌kJ‌mol−1
(b)
I=0.5At=2‌ hours ‌‌=2×60×60s=7200 s
Q‌=It
‌=0.5×7200
‌=3600‌ coulombs ‌
A flow of 96500c is equal to flow of 1 mole of electrons is 6.023×1023 electrons.
∴3600c is equivalent to of electrons
‌=‌

6.023×1023

96500

×3600
‌=2.246×1022‌ electrons ‌

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