Write reduction half-reactions for both of the given metals.
In a voltaic cell, each metal is immersed in a solution of its salt. Due to the flow of electrons in the circuit, the less electronegative metal will shed electrons and lose mass while the more electronegative metal will accept electrons and gain mass. However, since the cathode is the terminal of interest, write the reduction half-reaction of each of the given metals. (Include states of matter under the given conditions in your answer.)
Gold half-reaction:
Au^3+(aq) + 3e^- -> Au(s)
Aluminum half-reaction:
Al^3+(aq) + 3e^- -> Al(s)