[candidateof]

Hi,

If we have a net equation that is free of electrons, we just add or subtract the E^o for the Oxd and Red. But in the case of electrons exist, we calculate it from Gibbs energy. I was solving a question about a net equation with electrons and I have a problem. The question says:

Consider the following half rxn's:

1) (Cr₂O₇^2-) + 14H⁺ + 6e^--->2Cr³⁺ + 7H₂O    E₁^o=1.322 V

2)Cr³⁺ + 3e^- -->Cr   E₁^o=-0.744 V

What is the standard reduction potential for the half rxn from dichromate (Cr₂O₇^2-) to metallic chromium (Cr) in acetic medium? 

a)  0.289 V
b) -0.244 V
c)  0.988 V
d) -0.289 V
e)  0.244 V

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To obtain the reduction equation we add the 1^st to the 2^nd
G₁= -6 F E₁
G₂= -3 F E₂
G=-6*F*E₁-3*F*E₂
G=-9F(6/9(E^o₁)+3/9(E^o₂)
==> E^o=(6/9)(1.322V)+(3/9)(-0.744)= 0.633 V

Which is not one of the choices!

[candidateof]

I got an idea, if we multiply 2^nd half equation by 2 to cancel Cr³⁺ in the net reaction. We will have 6 e^- in the first and the second half equations, that yields 12 e^- in the net reaction:

G=-12*F((1/2)E₁^o+(1/2)E₁^o)
E=(1/2)(E₁^o)+(1/2)(E₂^o)=0.289 V

Is it right? If yes then we have to cancel Cr³⁺, why is that?