Consider an electrochemical cell based on the following cell diagram:

Pt | Pu3+(aq), Pu4+(aq) || Cl2(g), Cl−(aq) | Pt

Given that the standard cell emf is 0.35 V and that the standard reduction potential of chlorine is 1.36 V, what is the standard reduction potential E°(Pu4+/Pu3+), in volts? [Enter only a number.]

Cl is gaining elections which means it is the reduction half reaction. Reduction is associated with the Cathode.

E(cell) = E(cathode) - E(anode)

0.35v = 1.36v - E(anode)
solving for E(anode) gives you 1.01v

Pu+3 ==> Pu+4 + e Eo = ??

Cl2 + 2e ==> 2Cl^- Eo 1.36
============================
2Pu+3 + Cl2 ==> 2Pu+4 + 2Cl^- Ece.. = 0.35
What is ?? + 1.36 = 0.35
Solve for ?? which will be the oxidation potential (because that's the way the half cell is written above) so change the sign for the reduction problem (what the problem is asking).

Well, let's analyze the situation here, and by "analyze," I mean make fun of it.

So, we have Pu3+ and Pu4+, they sound like alien species from a sci-fi movie. And they are having a party with Cl2 and Cl-, like some sort of intergalactic mixer.

Now, the standard reduction potential of chlorine is 1.36 V. That's one electrifying party they've got going on there! But we need to find the standard reduction potential of Pu4+/Pu3+.

Hmm, let me think... If Cl2 has a higher standard reduction potential than our Pu4+/Pu3+ party, then I guess Pu4+/Pu3+ must be the weird, nerdy aliens in the corner of the party.

So, it's safe to say that the standard reduction potential of Pu4+/Pu3+ is lower than 1.36 V. But how much lower? Well, we know that the standard cell emf is 0.35 V. Let's assume that's the energy it takes for Pu4+/Pu3+ to summon the courage to join the party.

So, if we subtract 0.35 V from 1.36 V, we get... *drumroll*... 1.01 V! That's the standard reduction potential of Pu4+/Pu3+. They may be nerdy aliens, but they can still throw a decent party.

So, the answer is 1.01. Keep on partying, nerdy aliens!

To find the standard reduction potential E°(Pu4+/Pu3+), we can use the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential and the activities/concentrations of the species involved in the electrochemical cell.

The Nernst equation is given by:

E = E° - (RT/nF) * ln(Q)

Where:
- E is the cell potential
- E° is the standard cell potential
- R is the gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
- n is the number of electrons involved in the reaction
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient

In this case, the reaction quotient Q can be determined from the given cell diagram:

Pt | Pu3+(aq), Pu4+(aq) || Cl2(g), Cl−(aq) | Pt

Since the reactants are on the left side of the equation and the products are on the right side, Q is the ratio of the activity of the products over the activity of the reactants. The activities of solids and liquids are considered to be 1, so we focus on the species in solution.

Q = (activity of Cl−) / (activity of Pu3+)

Now, we can substitute the known values into the Nernst equation:

0.35 = E° - (RT/nF) * ln(Q)

We are given that the standard cell emf (E°) is 0.35 V, and the standard reduction potential of chlorine (Cl2/Cl−) is 1.36 V. Since the reaction involves a transfer of one electron, n = 1.

Plugging in the values:

0.35 = E° - (RT/F) * ln(Q)

(0.35) = (1.36) - (8.314 * T / 1 * 96485) * ln(Q)

Now, we need to rearrange the equation to solve for E°(Pu4+/Pu3+):

E°(Pu4+/Pu3+) = (8.314 * T / 96485) * ln(Q) + 1.36 - 0.35

Now, to find the value of E°(Pu4+/Pu3+), you need to know the temperature (T) and calculate the value of ln(Q) using the concentrations or activities of Cl− and Pu3+ in the electrochemical cell.

1.17 V