0:00
Solve the question.
00:01
So the formula to find the cell potential, which is e is equal to e naught, which is the standard cell potential minus r, which is the gas constant t, which is the temperature divided by n, which is the number of electrons transferred multiplied by the faraday's constant multiplied by the log of the reaction question, which is q.
00:32
So in this case, the reaction is 2 ag which is silver in the solid state reacts with 2 bromide ions plus cadmium 2 positive ions.
00:45
So this gives 2 moles of silver bromide in the solid state plus cadmium in the solid state.
00:54
So the balance half reactions will be ag positive plus 1 electron that gives ag.
01:02
So this is the reduction half reaction and the second half reaction will be cadmium 2 positive plus 2 electrons to give cadmium.
01:17
So this is the oxidation half reaction.
01:24
Now here the number of electrons transferred is 2 because 2 electrons are involved in both the half reactions.
01:34
Here the concentrations of kbr and cadmium nitrate is 0 .05 molar each.
01:49
So the reaction question can be calculated...