## Test 2, Question #7 [ENDORSED]

Hena Sihota 1L
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### Test 2, Question #7

What is the correct answer for this question: Calculate the standard potential for the following reaction using the standard reduction potentials on the last page of the test: Cr2O72-(aq)+14H+(aq)+12e-->2Cr(s)+7H2O(l)

Cam Bear 2F
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Joined: Thu Jul 27, 2017 3:01 am

### Re: Test 2, Question #7

This is similar to HW problem 14.27. So basically to get Cr2O72-(aq)+14H+(aq)+12e-->2Cr(s)+7H2O(l) from the reactions given on the last page of the test you have to combine the two reactions Cr2O72-(aq)+14H+(aq)+6e-->2Cr3+(aq)+7H2O(l) and Cr3+(aq)+3e--->Cr(s).
However $E^{\circ }$ isn't a state function so you can't just simply add up both of the $E^{\circ }$.
Instead, because Gibbs free energy is a state function, you convert $E^{\circ }$ of each half reaction to change in Gibbs free energy using $\Delta G^{\circ }=-nFE^{\circ }$.
First you have to multiply the second half reaction by 2 to get the right final reaction. Then you plug everything into the equation.
So $-nFE^{\circ }$ of the final reaction=$-nFE^{\circ }$
of the first reaction + 2 times $-nFE^{\circ }$ of the second reaction.

Anh Nguyen 2A
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### Re: Test 2, Question #7  [ENDORSED]

Cr2O72-(aq) +14H+(aq) +12e- -> 2Cr(s) +7H2O(l)
This reaction is made up of 2 reactions:
Cr2O72-(aq) +14H+(aq) +6e- -> 2Cr3+(aq) + 7H2O(l) E1=+1.33V
2Cr3+(aq) +6e- -> 2Cr(s) E2=-0.74V
ΔG=ΔG1+ΔG2
<=> -nFE = -n1FE1 - n2FE2
<=> -12FE = -6FE1 -6FE2
<=> E = 1/2E1 + 1/2E2
<=> E= 1/2x(1.33) + 1/2x-0.74 = 0.30V