## 6th edition 14.37

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

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### 6th edition 14.37

For part c, why is the standard potential of the reaction negative when the half reaction potentials were 0 and +1.36 and neither half reaction had to be flipped?
Additionally, when both pressures and concentrations are given, do you not have to convert the pressure into a concentration? In this question, the solutions manual only changed torr to atm but left it as atm when putting it into the Q equation.

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### Re: 6th edition 14.37

The cell diagram is anode left and cathode right and based on that you can calculate the cell potential (Cathode - anode). If you have both gas and a.q when writing Q, use concentration for a.q and bar for gas.

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