Determine whether or not each redox reaction occurs spontaneously in the forward direction.
a. Ni(s) + Zn²⁺(aq) → Ni²⁺(aq) + Zn(s)
b. Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s)
c. Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)
d. Pb(s) + Mn²⁺(aq) → Pb²⁺(aq) + Mn(s)

Question

Determine whether or not each redox reaction occurs spontaneously in the forward direction.

a. Ni(s) + Zn²⁺(aq) → Ni²⁺(aq) + Zn(s)

b. Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s)

c. Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)

d. Pb(s) + Mn²⁺(aq) → Pb²⁺(aq) + Mn(s)

Accepted Answer

hey everyone in this example, we're given the below redox reaction and we need to determine whether the forward direction of this reaction is going to be spontaneous. So because this is a redox reaction, we should recognize that we're gonna have two half reactions are first half reaction comes from our first reaction reactant which is 10 4 plus as a caddy on Where we produce other product, the 10 2 plus Catalon. Now we should recognize that we have a balanced number of 10 atoms on both sides of this reaction because we just have one atom of tin on both sides. However, on our reactant side we have a net charge of plus four, whereas on the product side we have a net charge of plus two. And so we we would want to go ahead and expand our reactive side to cancel out that net charge of plus four by adding two electrons. So that that would give us a net charge of plus two on both sides of our reaction. And so this means that our charges are balanced. But because we recognize that we added electrons to our reactant side for this half reaction, this would signal that this half reaction we would recall is going to occur as a reduction because again we added electrons to our reacting side. Now moving on to our second half reaction that's going to come from our second reactant which is our solid magnesium which according to our given reaction, we formed the product magnesium two plus Catalan. So our atoms of magnesium are balanced. However, we have a net charge of plus two on the product side of this half reaction. And so we would want to cancel this out so that we have a net charge of zero like we have on our reactant side here. So we would go ahead and cancel out this net charge of plus two on the product side by adding two electrons to our product side. And because we recall that we when we add electrons to our product side as we did here, this means that this half reaction occurs as an oxidation. Now we should recall next that our reduction half reaction occurs at the cathode of our voltaic cell. Whereas our oxidation half reaction occurs at the anodes of our voltaic cell. Our next step is to refer to our standard reduction potential table which we can find either online or in our textbooks and we would look up the oxidation of the 10 4 plus carry on. Which we would see according to this table has a cell potential equal to 0.15 volts. Whereas for our oxidation for our magnesium two plus carry on, we would see that that corresponds to a cell potential according to that table equal to a value of negative 2.37 volts. And so now that we have these values, we can go ahead and calculate our standard cell potential by recalling the falling formula where we take the cell potential of our cathode subtracted from the cell potential of our anodes And so plugging in our values from above, we would get the cell potential or our standard cell potential rather is equal to our self potential of our cathode which above we stated is equal to a value of positive 0.15V subtracted from our cell potential of our anodes, which above we stated as negative 2.37V. And so what we're going to get here is a value of positive 2.52V. And so we should recall that when our cell potential is greater than zero. This refers to our reaction being now spontaneous in the forward direction. And because we can agree that our self potential value which we calculated equal to positive 2.52V is greater than zero, we can say. Therefore our reaction is we can say yes, our reaction is spontaneous in the forward direction. And so for our final answer, we have determined that yes, our reaction is spontaneous in the forward direction. So 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

Determine whether or not each redox reaction occurs spontaneously in the forward direction.
a. Ni(s) + Zn²⁺(aq) → Ni²⁺(aq) + Zn(s)
b. Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s)
c. Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)
d. Pb(s) + Mn²⁺(aq) → Pb²⁺(aq) + Mn(s)

Verified Solution

Key Concepts

Redox Reactions

Standard Electrode Potentials

Spontaneity of Reactions

Determine whether or not each redox reaction occurs spontaneously in the forward direction. a. Ni(s) + Zn2+(aq) → Ni2+(aq) + Zn(s) b. Ni(s) + Pb2+(aq) → Ni2+(aq) + Pb(s) c. Al(s) + 3 Ag+(aq) → Al3+(aq) + 3 Ag(s) d. Pb(s) + Mn2+(aq) → Pb2+(aq) + Mn(s)

Verified Solution

Key Concepts

Redox Reactions

Standard Electrode Potentials

Spontaneity of Reactions

Determine whether or not each redox reaction occurs spontaneously in the forward direction.
a. Ni(s) + Zn²⁺(aq) → Ni²⁺(aq) + Zn(s)
b. Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s)
c. Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)
d. Pb(s) + Mn²⁺(aq) → Pb²⁺(aq) + Mn(s)