00:01
We are given these two half -reactions that take place within a galvanic cell, and we want to find the value for the change in gibbs -free energy reaction at standard conditions, as well as the equilibrium constant k.
00:14
We know that this is a galvanic cell, and so therefore the overall cell potential at standard conditions has to come out to a positive value.
00:23
So in order to rearrange these equations, in order to cancel out the total number of electrons on each side, while also giving a positive cell potential, we have to keep the reaction with the greater value for the standard reduction potential to be the reaction that undergoes reduction and reverse the other reaction to undergo oxidation.
00:48
And so that means that the second reaction, we have to reverse and also reverse the sign of the standard reduction potential.
00:58
And when we do that and combine the equations, we see that.
01:01
That we need to multiply the top equation by three, and the bottom equation by four, so we cancel off 12 total electrons on each side.
01:15
And so we are left with 3m4 plus aqueous plus 4n solid goes to 3m solid plus 4n3 plus aqueous and when we add the two potentials together to find the overall cell potential at standard conditions we see that it comes out to 0 .27 volts now we can use that value to solve for delta g at standard conditions so delta g is equal to negative n f e n we saw was 12 moles of electrons...