A positive E cell value indicates that the electrochemical reaction is favorable due to the relationship between Gibbs free energy and cell potential. The equation ΔG^0 = -nFE^0_cell shows that a negative free energy change corresponds to a positive E cell, making the reaction plausible. This is rooted in thermodynamic conventions where ΔG^0 < 0 suggests that the reaction will proceed spontaneously. Additionally, the relationship ΔG^0 = -RT lnK implies that a negative ΔG^0 leads to a reaction quotient K greater than 1, indicating product formation is favored. Understanding these conventions clarifies the implications of electrode potentials in electrochemical cells.
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Bladibla
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When we work out the E cell of with the electrode potential values of Zinc and copper (for example) Why does a positive E cell value indicate that the reaction is plausible?
Of course, now this translates to another convention : that a negative free energy change tells you that the reaction is plausible. To make this more intuitive, it may help to go a step further.
TL;DR Summary: cannot find out error in solution proposed.
[![question with rate laws][1]][1]
Now the rate law for the reaction (i.e reaction rate) can be written as:
$$ R= k[N_2O_5] $$
my main question is, WHAT is this reaction equal to?
what I mean here is, whether
$$k[N_2O_5]= -d[N_2O_5]/dt$$
or is it
$$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ?
The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...