(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.
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!
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