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What are the half-reactions for a galvanic cell with aluminum and gold electrodes?A.Al(s) Al3+(aq) + 3e- and Au+(aq) + e- Au(s)B.Al(s) Al3+(aq) + 3e- and Au(s) Au+(aq) + e-C.Al3+(aq) + 3e- Al(s) and Au+(aq) + e- Au(s)D.Al3+(aq) + 3e- Al(s) and Au(s) Au+(aq) + e-

Question

What are the half-reactions for a galvanic cell with aluminum and gold electrodes?A.Al(s) Al3+(aq) + 3e- and Au+(aq) + e- Au(s)B.Al(s) Al3+(aq) + 3e- and Au(s) Au+(aq) + e-C.Al3+(aq) + 3e- Al(s) and Au+(aq) + e- Au(s)D.Al3+(aq) + 3e- Al(s) and Au(s) Au+(aq) + e-

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Solution

The correct half-reactions for a galvanic cell with aluminum and gold electrodes are:

Al(s) → Al3+(aq) + 3e- and Au3+(aq) + 3e- → Au(s)

So, the correct option is:

Al(s) → Al3+(aq) + 3e- and Au3+(aq) + 3e- → Au(s)

This is because in a galvanic cell, the anode (Aluminum in this case) undergoes oxidation (loss of electrons), and the cathode (Gold in this case) undergoes reduction (gain of electrons).

This problem has been solved

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