Question

question 16 of 25 what are the half - reactions for a galvanic cell with aluminum and gold electrodes? a. al^{3 + }(aq)+3e^{-}\to al(s) and au(s)\to au^{ + }(aq)+e^{-} b. al^{3 + }(aq)+3e^{-}\to al(s) and au^{ + }(aq)+e^{-}\to au(s) c. al(s)\to al^{3 + }(aq)+3e^{-} and au^{ + }(aq)+e^{-}\to au(s) d. al(s)\to al^{3 + }(aq)+3e^{-} and au(s)\to au^{ + }(aq)+e^{-}

Explanation:

Step1: Determine oxidation and reduction

In a galvanic cell, oxidation occurs at the anode and reduction at the cathode. Aluminum is more reactive than gold, so aluminum will be oxidized and gold - ion will be reduced.

Step2: Write oxidation half - reaction

Oxidation is the loss of electrons. Aluminum metal ($Al(s)$) loses 3 electrons to form aluminum ions ($Al^{3 + }(aq)$). The oxidation half - reaction is $Al(s)\to Al^{3 + }(aq)+3e^-$.

Step3: Write reduction half - reaction

Reduction is the gain of electrons. Gold ions ($Au^{ + }(aq)$) gain 1 electron to form gold metal ($Au(s)$). The reduction half - reaction is $Au^{ + }(aq)+e^-\to Au(s)$.

Answer:

C. $Al(s)\to Al^{3 + }(aq)+3e^-$ and $Au^{ + }(aq)+e^-\to Au(s)$