Question with E=E' - 0.0592/n*logQ Plz Help!

[dbangell]

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I have a question regarding to the nonstandard cell potential.
The formula is E=E'-0.0592/n*logQ, but when shift to the right Ecell increases.
That troubles me that if Q increased, logQ would be bigger and minus a bigger number would be a decrease. Am I misunderstanding sth? Thanks.

 

[dbangell]

T_T anyone know the answer to the question?

 

[BrutalViking]

The log of a number smaller than one is a negative number. From what I understand, the reaction tries to reach equilibrium or Q tries to be like K. Since the reaction is shifting to the right, we want to make more products. In other words we have more reactants than products. Q is a ratio of [products]/[reactants]. If we have less products than reactants then Q is a fraction. Thus log of a fraction is a negative number. The minus sign after the E' in the formula would cancel the new "-" from the log and would change to a plus sign. Thus the formula would then be E=E'+(0.0592/n*logQ)

I hope that helped.

 

[dbangell]

The log of a number smaller than one is a negative number. From what I understand, the reaction tries to reach equilibrium or Q tries to be like K. Since the reaction is shifting to the right, we want to make more products. In other words we have more reactants than products. Q is a ratio of [products]/[reactants]. If we have less products than reactants then Q is a fraction. Thus log of a fraction is a negative number. The minus sign after the E' in the formula would cancel the new "-" from the log and would change to a plus sign. Thus the formula would then be E=E'+(0.0592/n*logQ)

I hope that helped.

Thx, now I understand