aamc9 ps #30 half cell

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Havik21

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Find standard emf for the galvanic cell in which the following occurs:
2Na(s) + Cl2(g) --> 2Na+ + 2Cl-

Half Rections:
Na+ + e ---> Na(s) Ereduction: -2.71
Cl2 +2e ---> + 2Cl- Ereduction: +1.36

The answer is 4.07V

My answer was 6.78V
Why is it not necessary to multiply the Reduction Potential of Na(s) by 2 as I did?
 
Find standard emf for the galvanic cell in which the following occurs:
2Na(s) + Cl2(g) --> 2Na+ + 2Cl-

Half Rections:
Na+ + e ---> Na(s) Ereduction: -2.71
Cl2 +2e ---> + 2Cl- Ereduction: +1.36

The answer is 4.07V

My answer was 6.78V
Why is it not necessary to multiply the Reduction Potential of Na(s) by 2 as I did?
You NEVER multiply the reduction potential even when balancing the equation...I dont remember why but I learned it from TBR.
 
Last edited:
Find standard emf for the galvanic cell in which the following occurs:
2Na(s) + Cl2(g) --> 2Na+ + 2Cl-

Half Rections:
Na+ + e ---> Na(s) Ereduction: -2.71
Cl2 +2e ---> + 2Cl- Ereduction: +1.36

The answer is 4.07V

My answer was 6.78V
Why is it not necessary to multiply the Reduction Potential of Na(s) by 2 as I did?

Yes, you don't multiply the E values when you double the half reactions. Only the signs change.
 
standard reduction/oxidation potentials are intrinsic quantities. They represent the energy PER unit charge. If you increase the amount of energy (by increasing coefficients) you also increase the charge involved such that the ratio remains constant.

Think of it as the attraction of a species to electrons. This doesn't change if you increase the amount of compound. You can even make a parallel to electronegativity. Does the electronegativity of oxygen increase if you have more oxygens in a compounds? Nah.
 
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