Q.

FeO42+2.0V Fe3+0.8V Fe2+0.5V Fe0

In the above diagram, the standard electrodepotentials are given in volts (over the arrow). The value of EFeO42/Fe2+Θ is

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a

2.1 V

b

1.7 V

c

1.4 V

d

1.2 V

answer is A.

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Detailed Solution

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ΔG4o=ΔG1o+ΔG2on4FE4o=n1FE10n2FE2o+4E4o=3×2+(1×0.8)E4o=6.84VE4o=1.7V

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FeO42−→+2.0V Fe3+→0.8V Fe2+→−0.5V Fe0In the above diagram, the standard electrodepotentials are given in volts (over the arrow). The value of EFeO42−/Fe2+Θ is