what is oxidized and what is reduced when aluminum reacts with aqueous base to produce alum and hydrogen gas?
Let V be the volume of the aluminium ping pongThen its mass will be 2.7 V gramWeight will be 2.7 gVEqual amount of liquid will be displacedIts mass will be V x, where x is the density of the liquid to be foundWeight is Vxg The apparent weight of the aluminium will be (2.7-x)VgThis is given in the problem as 70/100 2.7VgHence 2.7-x 0.72.7 cancelling g and V on both sidesSolving, x 0.32.7 0.81 g/cm^3 The density of the liquid is 0.81 g/cm^3 which is nothing but the kerosene.
2Al(s) + 2KOH(aq) + 6H2O(l) - 2KAl(OH)4(aq) + 3H2 (g) Oxidation is loss of electrons Reduction is gain of electrons Remember that electrons are the negative chargeex1 electron goes to 0 electrons (charge -1+ 1)is an oxidationAlthough the overall charge increases, all we are looking at is the electrons in the speciesAnything in it's natural state (bonded only to itself) has a charge of 0 Aluminum goes from a charge of 0 to +3 therefore it is oxidized Hydrodgen has (H2) is also it it's natural state, therefore it has a charge of 0Since Hydrogen in water has a charge of +1 it is gaining electrons to reach a charge of 0, therefore it is reducedTo support the identity of H2 you can capture a little gas produced in a test tube and light it on fireH2 is flammable so it should pop
