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 -
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.
E°cell = 0.41 V
Step 4: Calculate the equilibrium constant (K) using the Nernst Equation.
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 = ?
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Based on our data, we think this problem is relevant for Professor Rothberg's class at UR.
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.