KVPY SX SB Exam 20141 Question Paper

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Question : 108 of 120

Marks: +1, -0

The standard free energy change (in J) for the reaction

3Fe^{2+}(aq) + 2Cr(s) = 2Cr^{3+}(aq) + 3Fe(s)

given

E°_{Fe^{2+}\/Fe} =-0.44 V

and

E°_{Cr^{3+} \/Cr }= -0.74 V

is (F = 96500 C)

57,900
-57,900
-173,700
173,700

Solution:

3F ^{2+} + 2Cr \ { ––––▸} 2Cr^{3+} + 3Fe

E°_\cell = ( -0.44) + (0.74) = 0.3

volt
n = 6

∆Gº = -nFEº = - (6 × 96500 × 0.3)

J

= -173,700

J

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