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What are the half-reactions for a galvanic cell with aluminum and gold electrodes?
A. A3+ (aq) + 3e > Al(s) and Au+(aq) + e* → Au(s)
B. Al(s) → A13+(aq) + 3e and Aut(aq) + e
Au(s)
C. AB+(aq) + 3e – Al(s) and Au(s) — Au (aq) + e
D. Al(s) → AB+(aq) + 3e and Au(s) → Au(aq) + e

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Answer:B is right

Explanation: Al loses electrons and is oxidised, Au gains electron. And is reduced

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