For the following battery: Cd(s) | CdCl2(aq) || Cl–(aq) | Cl2(l) | C(s)

(a) Write the reduction half-reaction occuring at the C(s) electrode. (Include physical states of reactants and products.)

C(s) electrode:?

(c) Calculate the mass of Cl2 consumed if the battery delivers a constant current of 589 A for 82.0 min. ?

Please help! thank you!

(a) The reduction half-reaction occurring at the C(s) electrode is:

Cl2(l) + 2e- → 2Cl-(aq)

Note that the half-reaction is the reverse process of the one you provided. In the reduction reaction, Cl2 gas is being reduced at the C(s) electrode.

(c) To calculate the mass of Cl2 consumed, we first need to determine the total charge passed through the battery by multiplying the current and time. The time should be converted to seconds.

total charge (Coulombs) = current (Amperes) × time (seconds)
total charge = 589 A × 82.0 × 60 s = 2890920 C

Now, we can use the Faraday constant (F) which is the quantity of electrical charge (in Coulombs) required to reduce one mole of an ion or molecule. F is approximately 96,485 C/mol.

moles of Cl2 consumed = total charge(Faradays) / (moles electrons/Faraday) * (moles Cl2/moles electrons)
moles of Cl2 = 2890920 C / (96485 C/mol * 2)
moles of Cl2 = 14.995

Next, we can calculate the mass of Cl2 consumed using the molar mass of Cl2, which is 35.45 g/mol * 2.

mass of Cl2 = moles of Cl2 × molar mass of Cl2
mass of Cl2 = 14.995 mol × 70.90 g/mol = 1063.51 g

Thus, the mass of Cl2 consumed when the battery delivers a constant current of 589 A for 82.0 minutes is approximately 1063.51 grams.

(a) The reduction half-reaction occurring at the C(s) electrode is:

C(s) + 2Cl^-(aq) → Cl2(g) + 2e^-

(c) To calculate the mass of Cl2 consumed, we need to use Faraday's laws of electrolysis.

First, calculate the total charge passed through the circuit:

Q = I * t

where Q is the total charge (in coulombs), I is the current (in amperes), and t is the time (in seconds).

Given that the current is 589 A and the time is 82.0 min, convert the time to seconds:

t = 82.0 min * 60 s/min = 4920 s

Then calculate the total charge passed:

Q = 589 A * 4920 s = 2,895,480 C

Next, use Faraday's constant (F) to convert the charge to moles of electrons:

1 mol e^- = F coulombs

F = 96,485 C/mol e^-

So, the number of moles of electrons passed is:

moles of e^- = Q / F

moles of e^- = 2,895,480 C / 96,485 C/mol = 30 moles of e^-

Since the balanced chemical equation shows that 2 moles of electrons are required to produce 1 mole of Cl2, the number of moles of Cl2 consumed is half the number of moles of electrons:

moles of Cl2 = 30 moles of e^- / 2 = 15 moles of Cl2

Finally, use the molar mass of Cl2 to calculate the mass of Cl2 consumed:

molar mass of Cl2 = 35.45 g/mol

mass of Cl2 = moles of Cl2 * molar mass of Cl2

mass of Cl2 = 15 moles * 35.45 g/mol = 531.75 g

Therefore, the mass of Cl2 consumed if the battery delivers a constant current of 589 A for 82.0 min is 531.75 grams.

To determine the reduction half-reaction occurring at the C(s) electrode, we need to look at the overall cell reaction and identify the species that are being reduced at the C(s) electrode.

The overall cell reaction can be written as:
Cd(s) + Cl2(aq) -> CdCl2(aq) + Cl-(aq)

In this reaction, Cd(s) is being oxidized to CdCl2(aq), while Cl2(aq) is being reduced to Cl-(aq). Therefore, the reduction half-reaction at the C(s) electrode is the reduction of Cl2(aq) to Cl-(aq).

The reduction half-reaction is represented as:
2e- + 2Cl2(aq) -> 4Cl-(aq)
(Note: The physical states of the reactants and products should be mentioned.)

Now let's move on to calculating the mass of Cl2 consumed if the battery delivers a constant current of 589 A for 82.0 min.

To calculate the mass of Cl2 consumed, we need to use Faraday's law of electrolysis and the stoichiometry of the reaction.

Step 1: Calculate the total charge passed in Coulombs (C).
- The charge passed can be calculated using the formula: charge (C) = current (A) × time (s).
- Convert 82.0 min to seconds by multiplying it by 60: 82.0 min × 60 s/min = 4920 s.
- Total charge passed = 589 A × 4920 s = 2,897,680 C.

Step 2: Convert the charge passed to the number of moles of electrons.
- One mole of electrons is equal to 96,485 C (Faraday's constant).
- Moles of electrons = total charge passed (C) / Faraday's constant (C/mol).
- Moles of electrons = 2,897,680 C / 96,485 C/mol = 30 mol.

Step 3: Apply stoichiometry to convert moles of electrons to moles of Cl2.
- From the balanced equation, we know that 2 moles of electrons are required to produce 1 mole of Cl2.
- Moles of Cl2 = 30 mol / 2 = 15 mol.

Step 4: Convert moles of Cl2 to mass of Cl2.
- The molar mass of Cl2 is approximately 70.9 g/mol.
- Mass of Cl2 = moles of Cl2 × molar mass of Cl2.
- Mass of Cl2 = 15 mol × 70.9 g/mol = 1063.5 g.

Therefore, the mass of Cl2 consumed if the battery delivers a constant current of 589 A for 82.0 min is 1063.5 g.