Q.

If EFe+2/Fe0=x1V ; EFe+3/Fe+20=x2V,   EFe+3/Fe0 is

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a

2x1+x24

b

2x1+x23

c

2x1+x22

d

2x1+x2

answer is B.

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Detailed Solution

Fe+2+2e−→Fe; E10=x1 and ∆G10=-2FE10=-2Fx1 Fe+3+e−→Fe+2; E20=x2 and ∆G20= -1FE20= -1Fx2 Fe+3+3e−→Fe; E30=?; ∆G30=-3FE30=∆G10+∆G20 -3FE30=-F(2x1+x2) E30=2x1+x23
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If EFe+2/Fe0=x1V ; EFe+3/Fe+20=x2V,   EFe+3/Fe0 is