Answer:
A. cell potential.
Explanation:
Hello there!
In this case, according to the Nerst equation:
[tex]Ecell=E\°cell-\frac{0.0592}{n}log(Q)[/tex]
We can say that at equilibrium Ecell=0 and Q=K, so:
[tex]0=E\°cell-\frac{0.0592}{n}log(K)[/tex]
In such a way, we can solve for K as follows:
[tex]log(K)=\frac{nE\°cell}{0.0592}\\\\K=10^{\frac{nE\°cell}{0.0592}}[/tex]
Thus, we infer that E° in this equation is A. cell potential.
Regards!
