00:01
So we're asked to determine from cell notation what the balanced cell reactions are for a few different examples here.
00:12
For our first case we have silver being oxidized to silver cation at the anode and the cathode has a reduction of hydrogen cations to form hydrogen gas.
00:25
So let's first divide this into each of the balanced half reactions.
00:31
So at the anode we have silver becoming silver cation and producing a mole of electrons in the process.
00:40
That's our oxidation half reaction.
00:43
Our reduction half reaction then involves hydrogen cations becoming hydrogen gas in the presence of a platinum electrode.
00:53
So we need two hydrogen cations to balance our number of hydrogens on each side of the reaction.
01:04
But in order to balance charges we now have a plus two charge so two electrons are going to be needed in this reduction half reaction to completely balance it.
01:17
So to complete the process and have an overall balanced reaction the electrons gained which is two here must be equal to the electrons lost which is only one in our oxidation half reaction.
01:32
So we need to multiply that oxidation by a factor of two and this is then going to become two silvers, two silver cations and there are the two electrons.
01:44
Now that will balance the two electrons gained.
01:47
We'll add those two half reactions together to give us an overall balanced reaction here and cancel out the electrons lost and gained in the process.
02:00
So we have two silvers, two hydrogen cations producing hydrogen and two silver cations.
02:12
And again the electrons, two electrons on either side of the reaction cancel out in the process to give us our overall balanced reaction.
02:23
So our next example has dichromate and chromate reacting in the anode compartment and sorry that would be the cathode compartment and the bromide and bromine reacting in the anode compartment here.
02:47
So let's look at the overall balanced reactions here.
02:50
So first we have the dichromate being reduced to chromium 3 plus ion in an acidic environment.
02:59
And so we have seven oxygens we need to account for and so to complete that presence of seven oxygens we're going to go ahead and add seven waters to the product side.
03:17
And we also have an imbalance of chromiums.
03:23
We need two on each side so we're going to place a coefficient of two in front of our chromiums.
03:28
We now generated 14 hydrogens so to balance that properly we need to place 14 hydrogen cations on our reactant side as well.
03:39
And now all of our atoms balance properly but not our charges.
03:45
We have 14 plus minus 2 an overall plus 12 charge on the reactant side but 2 times 3 only a plus 6 charge on the product side.
04:02
So to fix that problem we're going to add 6 negatives or 6 electrons so that our charge balance works out properly as well...