Determine whether or not each redox reaction occurs spontaneously in the forward direction. b. 2 Ag+(aq) + Ni(s) ¡ 2 Ag(s) + Ni2+(aq)

Video transcript

hey everyone were given the below redox reaction. And we need to determine whether the four directions of this reaction is going to be spontaneous because we have a redox reaction. We should recognize that we're going to have two half reactions which come from our reactant. So for our first reactant we have barium two plus catalon which produces our product solid barium. We should recognize that we do have balance atoms. However, our net charge here is a plus two charge on the reactant side. Whereas on our product side we have a net neutral charge, meaning that we need to cancel out this plus two charge here on the reactive side by expanding our reactive side and adding two electrons to cancel out that plus two net charge. And so because we added electrons to our react inside, we would recognize that this half reaction occurs as a reduction. For our second half reaction, we have our second reactant which is our two moles of our sodium metal. And this is going to go ahead and form our product two moles of the carry on of sodium. And so what we should recognize is that we do have balanced atoms. We have two atoms of sodium on both sides. However, we have a net Charge of Plus one on the product side here and so on our reactive side, we need that to match with our net charge of zero. So we want to cancel out this net charge of we have a coefficient of two times plus one which would actually give us a charge of plus two, which we would cancel out by adding two electrons here on the product side. And because we added electrons to our product side, we would recognize that this half reaction occurs as an oxidation. So our next step is to recall that our reduction half reaction occurs at the cathode of our voltaic cell, whereas our oxidation half reaction will occur at the an ode of art voltaic cell. And so our next step is to recall our standard reduction potential table where we would see for the oxidation of our Barium two Plus Catalon, we're going to have a self potential value Equal to -2.90V at our cathode. Whereas for our oxidation of our sodium catalon according to this table, we're going to have a cell potential value Equal to negative 2.71V for our node. And so now we can go ahead and calculate our standard cell potential equal to our self potential of our cathode. Subtracted from our self potential of our an ode. So this is something that we should recall. And we're going to plug in our known values from above. So we would get that our standard cell potential value is equal to our self potential of our cathode which above we stated is equal to a value of negative 2.90V subtracted from our self potential of our node, which will plug in from above As -2.71V. And this is going to give us a standard cell potential value equal to negative 0.19V according to our calculators. And so what we should recall now is that when we have a standard cell potential value that is less than zero. This means that our reaction is going to be non spontaneous in the forward direction. And so because we can agree that our self potential value, Which we stated is equal to negative 0.19V is less than zero. We can say therefore our reaction, we can say therefore no, our reaction is not spontaneous, it's going to occur non spontaneous in the forward direction. And so for our final answer, we can confirm our self potential value here means that our reaction is going to be non spontaneous in the forward direction and just so that's visible, just scoot this over a bit. So I hope that everything I explained was clear. If you have any questions, please leave them down below. Otherwise, I'll see everyone in the next practice video