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PLEASE HELPPP WILL MARK BRAINLEST!!



PLEASE HELPPP WILL MARK BRAINLEST class=

Sagot :

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!