Given the following half-reactions and associated standard reduction potentials:
AuBr−4(aq)+3e−→Au(s)+4Br−(aq)
E∘red=−0.858V
Eu3+(aq)+e−→Eu2+(aq)
E∘red=−0.43V
IO−(aq)+H2O(l)+2e−→I−(aq)+2OH−(aq)
E∘red=+0.49V
Sn2+(aq)+2e−→Sn(s)
E∘red=−0.14V
Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf.
Express your answer as a chemical equation. Identify all of the phases in your answer.
I figured out that
Sn2+(aq)+2e−→Sn(s)
E∘red=−0.14V
-AND-
Eu3+(aq)+e−→Eu2+(aq)
E∘red=−0.43V
add up to be the smallest, but I am unsure how to go from there...
If that is the smallest Ecell, then turn those half cells into a reaction. Arrange the half cells so they produce a positive voltage.
2Eu^2+ + Sn^2+ ==> 2Eu^3+ + Sn
So I multiply Eu3+(aq)+e−→Eu2+(aq) by two so there is two e- in both half reactions. I also remember that I need to add H+ and OH- onto one or both sides depending if the reaction is acid or basic. How do I determine that?
The Eo voltage in tables may changes as a result of acid or basic conditions; however, there are no H or OH ions in the half reactions you've written so you need not worry about that.
Well, it seems like Tin (Sn) and Europium (Eu) are having a competition to see who can be less positive! Let's see what happens when they combine:
Sn2+(aq) + 2Eu3+(aq) + 2e− → Sn(s) + 2Eu2+(aq)
In this reaction, the Tin ions (Sn2+) are being reduced and gaining electrons, while the Europium ions (Eu3+) are being oxidized and losing electrons. The resulting products are solid Tin (Sn), and Eu2+ ions in solution.
Now, let's check the overall cell emf for this reaction:
E°cell = E°red(cathode) + E°red(anode)
= E°red(Eu2+) + E°red(Sn2+)
= -0.43 V + (-0.14 V)
= -0.57 V
So, the cell emf for this reaction is -0.57 V, making it the smallest positive emf.
Just be careful, even though it's the "smallest positive," it's still negative overall! Better luck next time, Tin and Europium!
To find the combination of half-cell reactions that leads to the smallest positive cell emf, we need to consider the reduction potentials of the given half-reactions. The smaller the reduction potential, the more likely the reaction will occur in the forward direction.
First, let's rewrite the reduction reactions by flipping the sign of the reduction potentials to represent oxidation reactions:
Au(s) → AuBr−4(aq) + 3e−
E∘ox = +0.858V
Eu2+(aq) → Eu3+(aq) + e−
E∘ox = +0.43V
I−(aq) + 2OH−(aq) → IO−(aq) + H2O(l) + 2e−
E∘ox = -0.49V
Sn(s) → Sn2+(aq) + 2e−
E∘ox = +0.14V
We want the smallest positive cell emf, which means we want to combine the reactions that have the smallest reduction potentials (or the largest oxidation potentials).
From the given half-reactions, Sn(s) → Sn2+(aq) + 2e− has the smallest oxidation potential of +0.14V, and Eu2+(aq) → Eu3+(aq) + e− has the second smallest oxidation potential of +0.43V.
To find the overall cell reaction, we need to cancel out the electrons on both sides of these reactions. Since Sn(s) needs 2 electrons while Eu2+(aq) provides only 1 electron, we need to multiply the Eu2+(aq) reaction by 2:
2 × (Eu2+(aq) → Eu3+(aq) + e−)
2Eu2+(aq) → 2Eu3+(aq) + 2e−
Now, add the two reactions together to obtain the overall cell reaction:
Sn(s) + 2Eu2+(aq) → Sn2+(aq) + 2Eu3+(aq)
The phases for the reactants and products are:
Sn(s) (solid), Eu2+(aq) (aqueous), Sn2+(aq) (aqueous), Eu3+(aq) (aqueous)
Therefore, the chemical equation for the combination of half-cell reactions that leads to the smallest positive cell emf is:
Sn(s) + 2Eu2+(aq) → Sn2+(aq) + 2Eu3+(aq)