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CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

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Question : 12 of 26
Marks: +1, -0
(a) The cell in which the following reaction occurs:
2Fe3+(aq)+2I−(aq)→2Fe2+(aq)2\mathrm{Fe}^{3+}(\mathrm{aq})+2\mathrm{I}^{-}(\mathrm{aq})\rightarrow2\mathrm{Fe}^{2+}(\mathrm{aq}) +I2(s)+\mathrm{I}_2(\mathrm{s})
has E  cell  ∘=0.236 VE^\circ_{\;\text{cell}\;}=0.236\,\mathrm{V} at 298 K298\,\mathrm{K}. Calculate the standard Gibb's energy of the cell reaction.
(Given : 1 F=96,500 C mol−11\,\mathrm{F}=96,500\,\mathrm{C}\,\mathrm{mol}^{-1} )
(b) How many electrons flow through a metallic wire if a current of 0.5 A0.5\,\mathrm{A} is passed for 2 hours ?
(Given : 1 F=96,500 C mol−11\,\mathrm{F}=96,500\,\mathrm{C}\,\mathrm{mol}^{-1} )
Solution:  
  (a)  \;\text{(a)}\;   2Fe3++2e−→2Fe2+\;2\mathrm{Fe}^{3+}+2\mathrm{e}^{-}\rightarrow2\mathrm{Fe}^{2+}
2I−→I2+2e−2\mathrm{I}^{-}\rightarrow\mathrm{I}_2+2\mathrm{e}^{-}
For the given cell reaction, n=2n=2.
ΔG∘=−nFE  cell  ∘\Delta G^\circ=-nFE^\circ_{\;\text{cell}\;}
  =−2×96500×0.236\;=-2 \times 96500 \times 0.236
  =−45548 J mol−1\;=-45548\,\mathrm{J}\,\mathrm{mol}^{-1}
  =−45.55 kJ mol−1\;=-45.55\,\mathrm{kJ}\,\mathrm{mol}^{-1}
(b)
I=0.5At=2  hours    =2×60×60  s=7200  sI=0.5^{A t=2\;\text{hours}\;\;=2\times60\times60\;\text{s}=7200\;\text{s}}
Q  =ItQ\;=I t
  =0.5×7200\;=0.5 \times 7200
  =3600  coulombs  \;=3600\;\text{coulombs}\;
A flow of 96500  c96500\;\mathrm{c} is equal to flow of 1 mole of electrons is 6.023×10236.023 \times 10^{23} electrons.
∴3600  c\therefore 3600\;\mathrm{c} is equivalent to of electrons
  =  6.023×102396500×3600\;=\;\frac{6.023 \times 10^{23}}{96500} \times 3600
  =2.246×1022  electrons  \;=2.246 \times 10^{22}\;\text{electrons}\;
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