Test #2 #7

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Natalie LeRaybaud 1G
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am

Test #2 #7

Postby Natalie LeRaybaud 1G » Thu Mar 15, 2018 2:07 pm

Did anyone get this problem correct and know how to solve it?

Calculate the standard potential for the following reaction using the standard reduction potentials on the last page of the test:
(Cr2O7)^2- (aq) + 14H+ (aq) + 12e- --> 2Cr (s) + 7H20 (l)

Ishan Saha 1L
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: Test #2 #7

Postby Ishan Saha 1L » Thu Mar 15, 2018 2:17 pm

Hi!
To answer solve this problem you had to first identify the two half reactions involved. They were (Cr2O7)^2- (aq) +14H+(aq) +6e- ==> 2(Cr)^3+ (aq) + 7(H2O)(l) with a standard cell potential of 1.33V and Cr^3+ +3e- ==> Cr(s) with a standard cell potential of -0.74V. THEN because cell potential is NOT a state function we had to convert both of these values to their Standard Gibbs free energy values using the equation . After solving for standard Gibbs Free Energy of both half reactions, we could use Hess's law to add them up, because Gibbs Free Energy IS a state function.

Srbui Azarapetian 2C
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Re: Test #2 #7

Postby Srbui Azarapetian 2C » Thu Mar 15, 2018 2:22 pm

The two equations you use are
1: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ +7H2O . E=1.33V
2: Cr3+ + 3e- -> Cr(s) . E=-.74V

Notice that a sum of these after balancing the charges creates the overall equation. Since electric potential is not a state function, you have to use Gibbs Free Energy (delta G=-nFE)

G(overall)= G(equation 1) + G(equation 2)
-nFE= -nFE (1) + -nFE(2)
The Faradays Constants all cancel leaving you with:
-nE= -nE (1) + -nE(2)
The n is the moles of electron in each equation, so
-6E= -6(1.33V)+ -3(-.74V)
E=0.96V

Grace Han 2K
Posts: 33
Joined: Tue Nov 15, 2016 3:00 am

Re: Test #2 #7

Postby Grace Han 2K » Thu Mar 15, 2018 4:39 pm

Im confused as to where n(overall)=6. Can you please explain?

Katelyn B 2E
Posts: 29
Joined: Fri Sep 29, 2017 7:05 am

Re: Test #2 #7

Postby Katelyn B 2E » Thu Mar 15, 2018 8:01 pm

Another way to do this problem is by plugging in the values of E°=1.33 V and E°=-0.74 V into the equation ΔG°= -nFE° to solve for their respective ΔG° values. You add these values together to get the total ΔG°, then plug that value back into the ΔG°= -nFE° equation again and solve E° this time to get the standard potential. So, you essentially use the ΔG°= -nFE° equation three times.

Rachel N 1I
Posts: 48
Joined: Thu Jul 27, 2017 3:00 am

Re: Test #2 #7

Postby Rachel N 1I » Sat Mar 17, 2018 8:11 pm

Should the overall n = 12, not 6?


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