[AAMC 7] What is the electrode potential of the reaction

[sc4s2cg]

Given:

Cu^2+ + 2e ---> Cu E = +0.34V
2H2O ---> O2 + 4H^+ + 4e E = -1.23V

What is Ecell for the reaction below?
2Cu^2+ + 2H2O ---> 2Cu + O2 + 4H^+

A) -0.89V
B) +0.55V
C) +1.57V
D) +1.91V

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So I did it the way I knew how. I multiplied the top equation by two, then added the two electrode potentials together to get 0.55. The answer was -0.89V.

Is there a reason that we don't multiply E the same way we would deltaH when calculating enthalpy?

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[BB8730]

Is there a reason that we don't multiply E the same way we would deltaH when calculating enthalpy?

Yep.

E = volt = joules/coulombs

When you multiply the half reactions by stoichiometric factors for the purpose of balancing, both the energy and the charge increase by that same amount, leaving E unchanged.

deltaH is just joules, so you must multiply that by the stoichiometric factor to keep things even.

 

[Jepstein30]

Given:

Cu^2+ + 2e ---> Cu E = +0.34V
2H2O ---> O2 + 4H^+ + 4e E = -1.23V

What is Ecell for the reaction below?
2Cu^2+ + 2H2O ---> 2Cu + O2 + 4H^+

A) -0.89V
B) +0.55V
C) +1.57V
D) +1.91V

-----------------------------------------------
So I did it the way I knew how. I multiplied the top equation by two, then added the two electrode potentials together to get 0.55. The answer was -0.89V.

Is there a reason that we don't multiply E the same way we would deltaH when calculating enthalpy?

Its an intensive property, doesn't matter on how much you have. You can have 1 mol of Cu being reduced or 10,000 mols of Cu being reduced.. E of the reaction is still .34V.

So to find E(cell) just make sure the half equations match the order (as in, you have whats reduced and oxidized in the right places), make sure to flip the sign of any E you need to and then add them together.

In this case, they are both given as they appear. So just add 0.34 + -1.23 = -0.89.

 

[sc4s2cg]

Ah, I see.

So if I multiply the equation by 2, then we get

E = 2Joules/2Coulombs = E
or
H = 2J

So the electron potential's units cancel out to be the original, while the enthalpy is multiplied by two?

 

[BB8730]

So the electron potential's units cancel out to be the original, while the enthalpy is multiplied by two?

Yep, so you never have to multiply the potential by anything but you still need to multiply deltaH by the stoichiometric factor if there is one.