## Lecture Example

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Mia Navarro 1D
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### Lecture Example

During lecture today, Dr. Lavelle solved the potential energy of the cell given:
Ag+ (1.0 M) —> Ag+(0.1 M)
He then solved the potential out to be
E(cell) = Eº - (0.0592/n)log[P]/[R] = 0 - (0.0592/1)log10 = 0.0592 V

Maybe I miswrote this, but how is E(cell) positive in the end?

Anna Okabe
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: Lecture Example

E cell is positive because there is a log (0.1/1.0), which equals -1.
If you solve the equation, this -1 cancels out the negative in the equation, making it positive.

Michelle Lee 2E
Posts: 64
Joined: Thu Jul 27, 2017 3:01 am

### Re: Lecture Example

You can also reason out that since the product has a lower concentration than the reactant, more product will want to form which makes the reaction spontaneous. A spontaneous reaction has a positive E value.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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