A particular Down's cell operates for 1.00 × 10^5 s at a current of 96.5 A - VCE - SSCE Chemistry - Question 18 - 2002 - Paper 1
Question 18
A particular Down's cell operates for 1.00 × 10^5 s at a current of 96.5 A.
The amount of chlorine produced, in mole, is
A. 0.050
B. 5.00
C. 10.0
D. 20.0
Worked Solution & Example Answer:A particular Down's cell operates for 1.00 × 10^5 s at a current of 96.5 A - VCE - SSCE Chemistry - Question 18 - 2002 - Paper 1
Step 1
Calculate the Charge (Q)
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Answer
To calculate the total charge passed through the cell, we use the formula: Q=Iimest
where:
Q is the charge in coulombs,
I is the current in amperes (A),
t is the time in seconds (s).
Thus, Q=96.5extAimes1.00imes105exts=9.65imes106extC
Step 2
Determine Moles of Electrons (n)
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Answer
We can find the moles of electrons using the relation: n = rac{Q}{F}
where:
n is the number of moles,
F is Faraday's constant, approximately 96485 C/mol.
Substituting the charge: n = rac{9.65 imes 10^6 ext{ C}}{96485 ext{ C/mol}} = 100 ext{ mol e}
Step 3
Calculate Moles of Chlorine Produced
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Answer
According to the electrochemical reaction for chlorine production:
ightarrow 2Cl^- $$
This shows that 1 mole of $ Cl_2 $ is produced by 2 moles of electrons.
Thus, the moles of chlorine produced are:
$$ ext{Moles of } Cl_2 = rac{100 ext{ mol e}}{2} = 50 ext{ mol Cl}_2 $$
Step 4
Final Answer
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Answer
Given the options available, the closest answer is: B. 5.00
