Calculate E°_cell for each balanced redox reaction and determine if the reaction is spontaneous as written. b. MnO₂(aq) + 4 H⁺(aq) + Zn(s) → Mn²⁺(aq) + 2H₂O(l) + Zn²⁺(aq) c. Cl₂(g) + 2 F^–(aq) → F₂(g) + 2 Cl^–(aq)

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hey everyone in this example, we're given the below redox reaction and we need to calculate the standard cell potential as well as whether this cell potential is going to be spontaneous or non spontaneous for our following reaction. So we should recognize that this given reaction is a redox reaction, meaning we need to write out our two half reactions and that's going to come from our reactant. So our first reactant is the chromium three plus catty on which produces as a product, solid chromium. Now we have one atom of chromium on both sides of this half reaction. However, looking at our net charge, we have a net charge of plus three on the reactant side and a net charge of zero for our neutral atom of solid chromium. And we want to go ahead and cancel out this net plus three charge on the reactant side. So in order to do so we're going to expand our react inside and add three electrons here and this will cancel out this net plus three charge that we have a net charge of zero on both sides of our reaction. And so because we added electrons to our react inside, we should recognize that this half reaction occurs as a reduction. Now moving on to our second half reaction which comes from our second reactant, that's going to be our solid iron according to our given reaction, which as a product we produce one mole of the iron, three plus cat ion. Now our atoms are balanced, We have one atom of iron here and one atom of iron on a react inside. However, our net charge of plus three on the product side does not match our net charge of zero on the react inside. And so we need to cancel out this net charge of plus three. And so as before we're going to add three electrons. But now on the product side for the second half reaction. And because we added electrons on our product side, we can say that therefore the second half reaction is going to occur as an oxidation. Now, what we should recall next now that we've labeled each of our half reactions as reduction and oxidation. Is that our reduction half reaction occurs in our voltaic cell at the cathode. Where are oxidation? Half reaction occurs in our voltaic cell at the node. And so our next step is to call our standard reduction potential table. So we would find this table either online or in our textbooks and we're going to find for the oxidation of the iron three plus cat ion. We would see that it has a self potential value for an ode equal to negative 0.036V. And then also for our cathode where we have the oxidation of our chromium three plus carry on. We're going to see that that has a self potential value according to the table equal to negative 0.73V. And so now what we should recall is to calculate our standard cell potential. We're going to recall that That is equal to our self potential of our cathode. Subtracted from our cell potential of our an ode. And so plugging in our values above we would get that our standard cell potential is equal to Our value for our cathode, which above we stated is negative 0.73V Subtracted from our value for a node, which above we stated as a value of negative 0.036V. And so in our calculators, this is going to give us a value for our standard cell potential equal to negative 0.6 94 sorry, negative 0.6 94 volts. And so next we want to recall that when our self potential is less than zero, this would therefore mean that our reaction is non spontaneous in the forward direction. And so because we can say, we can agree that our self potential That we calculated above equal to negative 0.694V is less than zero. We can say therefore our reaction is non spontaneous in the forward direction. And so for our final answers, we would agree that we've determined our self potential here as well as the fact that our reaction is going to be non spontaneous in the four directions based on this value. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. Otherwise, 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. b. MnO₂(aq) + 4 H⁺(aq) + Zn(s) → Mn²⁺(aq) + 2H₂O(l) + Zn²⁺(aq) c. Cl₂(g) + 2 F^–(aq) → F₂(g) + 2 Cl^–(aq)

Verified Solution

Key Concepts

Standard Electrode Potential (Ec°ell)

Balancing Redox Reactions

Spontaneity of Reactions

Calculate E°cell for each balanced redox reaction and determine if the reaction is spontaneous as written. b. MnO2(aq) + 4 H+(aq) + Zn(s) → Mn2+(aq) + 2H2O(l) + Zn2+(aq) c. Cl2(g) + 2 F–(aq) → F2(g) + 2 Cl–(aq)

Verified Solution

Key Concepts

Standard Electrode Potential (Ec°ell)

Balancing Redox Reactions

Spontaneity of Reactions

Calculate E°_cell for each balanced redox reaction and determine if the reaction is spontaneous as written. b. MnO₂(aq) + 4 H⁺(aq) + Zn(s) → Mn²⁺(aq) + 2H₂O(l) + Zn²⁺(aq) c. Cl₂(g) + 2 F^–(aq) → F₂(g) + 2 Cl^–(aq)