# Problem: For the following half-reaction, ε° = —2.07 V:AlF63-(aq) + 3e - → Al(s) + 6F -(aq)Using data from the following table, calculate the equilibrium constant at 25°C for the reactionAl 3+(aq) + 6F -(aq) ⇌ AlF63-(aq)    K = ?

🤓 Based on our data, we think this question is relevant for Professor Rothberg's class at UR.

###### FREE Expert Solution

Overall reaction:     Al3+(aq) + 6F -(aq) ⇌ AlF63-(aq)

Step 1: Identify the anode and the cathode in the reaction.

AlF63-(aq) + 3e → Al(s) + 6F -(aq)                      E° = 2.07 V               ↓ E° → oxidation → anode

(since we also need AlF63-(aq)  on the product side)

Al3+(aq) + 3 e- Al(s)                                          E° = – 1.66 V               ↑ E° → reduction → cathode

Step 2: Write the overall reaction and determine the number of electrons transferred.

lose electrons oxidation → anode
gain electrons reduction → cathode

Al(s) + 6 F -(aq)  → AlF63-(aq) + 3e -
Al3+(aq) + 3 e- Al(s)
___________________________
Al3+(aq) + 6 F -(aq)  → AlF63-(aq)

# of electrons transferred = 3 e­-

Step 3: Calculate the cell potential of the reaction.

cell = 0.41 V

Step 4: Calculate the equilibrium constant (K) using the Nernst Equation. ###### Problem Details

For the following half-reaction, ε° = —2.07 V:

AlF63-(aq) + 3e → Al(s) + 6F -(aq)

Using data from the following table, calculate the equilibrium constant at 25°C for the reaction

Al 3+(aq) + 6F -(aq) ⇌ AlF63-(aq)    K = ? 