Calculate E°_cell for each balanced redox reaction and determine if the reaction is spontaneous as written. a. O₂(g) + 2 H₂O(l) + 4 Ag(s) → 4 OH^–(aq) + 4 Ag⁺(aq) b. Br₂(l) + 2 I^–(aq) → 2 Br^–(aq) + I₂(s)

Question

Accepted Answer

Hey everyone in this example, we're given the below bounced redox reaction and we need to calculate the standard cell potential of this reaction. Were also asked whether the forward direction of the redox reaction is spontaneous or whether it's non spontaneous. So what we should recognize is based on our given reaction. We have a redox reaction, meaning we need to write out our half reactions and that is going to come from our reactant. So looking at our first reactant, we have led to plus this Catalan produces the product solid lead and we should recognize that first, whether our atoms are balanced. So we have one atom of lead on the product side and one atom of lead on the reactive side. So our lead atom is balanced. However, we do not have a net balanced charge here. We have a net charge of plus two on the reactant side and we have a net charge of zero on the product side. So we want to cancel out this plus two charge here and we would do so by adding two electrons to our reactant side. So we can fill it in as plus two electrons here to give us a net charge of zero on both sides of our reaction. Now, because we recognize that we added electrons on our reactant side, we should recognize that this means that this half reaction occurs as a reduction and so now let's move on to our second half reaction, which comes from our second reactant which is our hydrogen sulfide gas According to the reaction. We form as a product solid sulfur and two moles of H. Plus. Now we can recognize that our hydrogen atoms are balanced. We have two on both sides of the reaction as well as our sulfur atoms are balanced. So our next step is to recognize whether the net charge is balanced on our reactive side we have a net charge of zero because we have a neutral compound. But on our product side we have a net charge of our coefficient of two added to this or multiplied by this plus one charge, which would give us a net charge of plus two on the product side. So we need to cancel out this net charge of plus two. And we're going to do so by adding two electrons to our product side. And because we recognize that we added two electrons to our product side. This half reaction is going to occur as an oxidation. Our next step is to recall that our half reaction that occurs as a reduction is going to occur in our electoral voltaic cell at the cathode. Whereas our oxidation half reaction occurs in our electoral voltaic cell at the node. And so at this point we want to recall our standard reduction potential table And we want to go and find our self potential value for our cathode which involves our lead to Pluskat ion oxidation. And we would see that according to this table, we're going to have a value equal to negative 0.13V. So this was for our cathode. And then we want to look up our cell potential for the oxidation of sulfur. And we would see this on our table that this has a cell potential for our anodes. Where sulfur is involved. Equal to a value of negative or sorry, positive zero .14V. So all of this information comes from this table. And now that we have this outlined, we can go ahead and recall our formula to calculate our standard cell potential and that is going to be equal to our self potential of our cathode, subtracted from the cell potential of our A node. So, plugging in what we know from our table above, we would see that our cell potential for our cathode we stated is equal to a value of negative 0.13V. And then this is subtracted from our cell potential of our anodes, which above we stated is a value of 0.14V. And so for our standard cell potential, we're going to get a value equal to negative 0.27V. So our next step is to recall that our cell potential when it is less than zero. This will mean that therefore our reaction is non spontaneous in the forward direction. And so because above we calculated that our standard cell potential is equal to a value of negative 0.27V. We can say that this is less than zero and so therefore our reaction is non spontaneous. So for our final answers, we can confirm that we've calculated our standard cell potential as negative 0.27V, and we can confirm that our reaction is therefore going to be non spontaneous based on this cell potential value. So everything highlighted in yellow represents our final answers. I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Calculate E°_cell for each balanced redox reaction and determine if the reaction is spontaneous as written. a. O₂(g) + 2 H₂O(l) + 4 Ag(s) → 4 OH^–(aq) + 4 Ag⁺(aq) b. Br₂(l) + 2 I^–(aq) → 2 Br^–(aq) + I₂(s)

Verified Solution

Key Concepts

Standard Electrode Potential (Ec°ell)

Redox Reactions

Spontaneity of Reactions

Calculate E°cell for each balanced redox reaction and determine if the reaction is spontaneous as written. a. O2(g) + 2 H2O(l) + 4 Ag(s) → 4 OH–(aq) + 4 Ag+(aq) b. Br2(l) + 2 I–(aq) → 2 Br–(aq) + I2(s)

Verified Solution

Key Concepts

Standard Electrode Potential (Ec°ell)

Redox Reactions

Spontaneity of Reactions

Calculate E°_cell for each balanced redox reaction and determine if the reaction is spontaneous as written. a. O₂(g) + 2 H₂O(l) + 4 Ag(s) → 4 OH^–(aq) + 4 Ag⁺(aq) b. Br₂(l) + 2 I^–(aq) → 2 Br^–(aq) + I₂(s)