The important thing here is to understand the Standard Electrode Potential for each of our metals involved. Whichever of the metals has a larger negative number is the one that will be our anode, the other will be our cathode. Therefore, for this reaction, we know that Aluminum is going to be our anode. With this reasoning, we know that where the anode is, so will oxidation occur. If we think about it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and feeding it to the other side. Since we know the salt bridge is there to balance out the lost electrons from the oxidation side, we know that the NO3- are going to flow to the left hand side to make up for the electrons that are going from the left hand side to the right hand side. Your salt bridge sends anions to the left hand side! Hope this helps! Solution The important thing here is to understand the Standard Electrode Potential for each of our metals involved. Whichever of the metals has a larger negative number is the one that will be our anode, the other will be our cathode. Therefore, for this reaction, we know that Aluminum is going to be our anode. With this reasoning, we know that where the anode is, so will oxidation occur. If we think about it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and feeding it to the other side. Since we know the salt bridge is there to balance out the lost electrons from the oxidation side, we know that the NO3- are going to flow to the left hand side to make up for the electrons that are going from the left hand side to the right hand side. Your salt bridge sends anions to the left hand side! Hope this helps!.