[rjbrivera]

Calculate the voltage for the following cell (pKsp) for AgBr=12.30; e^o Ag^+/Ag= 0.799 V). Identify the anode and cathode of the following cell. Identify the S.H.E. What is the purpose of the S.H.E. ? State the direction in which the electrons flow. State whether the cell is spontaneous or not spontaneous.

For the voltage I got 11.501.
I need help on the rest.

RJ

[Borek]

Show what you did and how.

11 volts is a nonsensical value, there are no cells with voltages higher (lower) than about 3-4 volts.

[rjbrivera]

12.30-0.799v= 11.501

[SinkingTako]

Hmm for the voltage of the AgBr/Ag⁺ why did you just take the voltage as Ksp??

Read up on the Nernst Equation (which I think you should have learnt or else your prof won't ask this question?).

Anyways, Nernst Equation states that for the cell Ag⁺ + e^- Ag that is non-standard (ie in this case, the concentration of Ag⁺ is not 1M, due to the presence of Br^-)
[tex]E=E^{o}-\frac{RT}{nF}ln{\frac{1}{[Ag^+]}}[/tex]

From the Ksp, you should be able to find the [Ag⁺] in solution. Then you know that the standard hydrogen cell has a potential of 0V, so can you solve the voltage of the cell?