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Chemistry Major
The University of Tanjungpura Pontianak, West Borneo, Indonesia
Q&A Practice Exam
1. An electrolysis experiment of 0.01 mol of chromium in an unknown salt was deposited at
the cathode when 0.04 moles of electrons were passed through an electrolyte solution
containing chromium. Determine the oxidation number of the chromium.
Solution
Calculate the number of moles of electrons that passed through the cell during electrolysis
1 mol Cr =
0.04
0.01
= 4 mol e
-
This means four moles of electrons are consumed on every mol chromium metal produced
in reaction at the cathode.
Thus, the reaction at the cathode involves reduction of Cr
4+
ions to chromium metal
Cr
4+
+ 4e
→ Cr
In conclution, the oxidation number of chromium must be +4
2. In the electrolysis of molten Al
2
O
3
using an inert electrode, calculate the volume of gas
produced (measured at standard temperature and pressure - STP) when a current of 12
amps is passed through the electrolyte for 100 minutes.
Solution
Calculate the number of coulomb (Q):
Q = I × t = 12 amps × 100 minutes × 60 seconds/minute
Q = I × t = 72000 coulombs
Al
2
O
3
2Al
3+
+ 3O
2
Oxidation in anode produced O
2
:
2O
2
(l) → O
2
(g) + 4e
1 mol of electrons = 1 faraday = 96500 coulombs
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That means for every mol of O
2
are produced from the reaction, four moles of electrons
are also produced.
The number of moles O
2
from the reaction:
4 𝐹
1 𝑚𝑜𝑙 𝑜𝑥𝑦𝑔𝑒𝑛
=
0,75 𝐹
𝑋 𝑚𝑜𝑙 𝑜𝑥𝑦𝑔𝑒𝑛
X mol O
2
=
0,75 𝐹 𝑥 1 𝑚𝑜𝑙 𝑜𝑥𝑦𝑔𝑒𝑛
4 𝐹
x mol O
2
= 0,1875 mol
In STP condition 1 mol of gas = 22.4 dm
3
Thus, volume of O
2
at STP = (0.1875 mol × 22.4 dm
3
/mol) = 4.2 dm
3
3. Diluted sulfuric acid solution is electrolyzed, calculate the volume of oxygen produced
when 6 moles of electrons are passed from an electrode to the other electrode. The volume
of oxygen is measured at room temperature and pressure (RTP) when every 1 mol of gas
is equal to 24 dm
3
.
Solution
Balance equation for the reaction of sulfuric acid solution:
H
2
SO
4(aq)
2H
+
(aq)
+ SO
4
2-
(aq)
Cathode : 2H
+
(aq)
+ 2e
-
(aq)
H
2(aq)
Anode : 2H
2
O
(l)
→ 4H
+
(aq)
+ O
2(g)
+ 4e
(aq)
The gas of O
2
is produced at anode.
Based on coefficient comparison of e
-
and O
2
:
mol of O
2
=
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
× moles of electrons
= ¼ × 6 = 1.5 mollah m
Volume of O
2
at RTP = 1.5 mol × 24 dm
3
/mol = 36 dm
3
4. Silver nitrate solution is electrolyzed, calculate the number of electrons needed to
precipitate 2.16 gram of silver at the cathode. (Ar Ag = 108 and L = Avogadro's constant)
Solution
Balance equation for the reaction of sulfuric acid solution:
AgNO
3(aq)
Ag
+
(aq)
+ NO
3
-
(aq)
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Cathode : Ag
+
(aq)
+ e
-
Ag
(s)
Anode : 2H
2
O
(l)
→ 4H
+
(aq)
+ O
2(g)
+ 4e
(aq)
The deposit of Ag is produced at cathode.
Based on coefficient comparison of e
-
and Ag, mol of e
-
= mol of Ag
The number of mol e
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴𝑔
𝐴𝑟 𝐴𝑔
=
2.16 𝑔𝑟𝑎𝑚
108
= 0.02 moles of electrons
The number of electrons = moles of electrons × L = 0.02 × L = L/200
5. Calculate the mass of chromium metal produced by electrolysis of K2Cr2O7 for 2 hours if
you use a current of 25 amps. (Ar Cr = 52)
Solution
2 hours = 2 hours × 60 minutes/hour × 60 seconds/minute = 7200 seconds
Oxidation number of Cr in K
2
Cr
2
O
7
is +6
Cr
+6
+ 6e
-
Cr
This means 1 mol of Cr
+6
needs 6 moles of electrons (6 faradays)
Mass of Cr =
𝐴𝑟 𝑥 𝐼 𝑥 𝑡
𝑖𝑜𝑛𝑠 𝑥 96500
=
52 𝑥 25 𝑥 7200
6 𝑥 96500
= 16.1658 gram

Unformatted Attachment Preview

Basic Chemistry – Electrolysis Calculations (Faraday) Chemistry Major The University of Tanjungpura Pontianak, West Borneo, Indonesia Q&A Practice Exam 1. An electrolysis experiment of 0.01 mol of chromium in an unknown salt was deposited at the cathode when 0.04 moles of electrons were passed through an electrolyte solution containing chromium. Determine the oxidation number of the chromium. Solution Calculate the number of moles of electrons that passed through the cell during electrolysis 0.04 1 mol Cr = 0.01 = 4 mol eThis means four moles of electrons are consumed on every mol chromium metal produced in reaction at the cathode. Thus, the reaction at the cathode involves reduction of Cr4+ ions to chromium metal Cr4+ + 4e– → Cr In conclution, the oxidation number of chromium must be +4 2. In the electrolysis of molten Al2O3 using an inert electrode, calculate the volume of gas produced (measured at standard temperature and pressure - STP) when a current of 12 amps is passed through the electrolyte for 100 minutes. Solution Calculate the number of coulomb (Q): Q = I × t = 12 amps × 100 minutes × 60 seconds/minute Q = I × t = 72000 coulombs Calculate the number of Faraday: 72000 coulombs : 96500 coulombs/faraday = 0.75 faradays Al2O3 → 2Al3+ + 3O2– Oxidation in anode produced O2: 2O2–(l) → O2(g) + 4e– 1 mol of electrons = 1 faraday = 96500 coulombs That means for every mol of O2 are produced from the reaction, four moles of electrons are also produced. The number of moles O2 from the reaction: 4𝐹 0,75 𝐹 = 1 𝑚𝑜𝑙 𝑜𝑥𝑦𝑔𝑒𝑛 𝑋 𝑚𝑜𝑙 𝑜𝑥𝑦𝑔𝑒𝑛 X mol O2 = 0,75 𝐹 𝑥 1 𝑚𝑜𝑙 𝑜𝑥𝑦𝑔𝑒𝑛 4𝐹 x mol O2 = 0,1875 mol In STP condition 1 mol of gas = 22.4 dm3 Thus, volume of O2 at STP = (0.1875 mol × 22.4 dm3/mol) = 4.2 dm3 3. Diluted sulfuric acid solution is electrolyzed, calculate the volume of oxygen produced when 6 moles of electrons are passed from an electrode to the other electrode. The volume of oxygen is measured at room temperature and pressure (RTP) when every 1 mol of gas is equal to 24 dm3. Solution Balance equation for the reaction of sulfuric acid solution: H2SO4(aq) → 2H+(aq) + SO42- (aq) Cathode : 2H+(aq) + 2e-(aq) → H2(aq) Anode : 2H2O(l) → 4H+(aq) + O2(g) + 4e–(aq) The gas of O2 is produced at anode. Based on coefficient comparison of e- and O2: mol of O2 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 × moles of electrons = ¼ × 6 = 1.5 mollah m Volume of O2 at RTP = 1.5 mol × 24 dm3/mol = 36 dm3 4. Silver nitrate solution is electrolyzed, calculate the number of electrons needed to precipitate 2.16 gram of silver at the cathode. (Ar Ag = 108 and L = Avogadro's constant) Solution Balance equation for the reaction of sulfuric acid solution: AgNO3(aq) → Ag+(aq) + NO3- (aq) Cathode : Ag+(aq) + e- → Ag(s) Anode : 2H2O(l) → 4H+(aq) + O2(g) + 4e–(aq) The deposit of Ag is produced at cathode. Based on coefficient comparison of e- and Ag, mol of e- = mol of Ag The number of mol e– = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴𝑔 𝐴𝑟 𝐴𝑔 = 2.16 𝑔𝑟𝑎𝑚 108 = 0.02 moles of electrons The number of electrons = moles of electrons × L = 0.02 × L = L/200 5. Calculate the mass of chromium metal produced by electrolysis of K2Cr2O7 for 2 hours if you use a current of 25 amps. (Ar Cr = 52) Solution 2 hours = 2 hours × 60 minutes/hour × 60 seconds/minute = 7200 seconds Oxidation number of Cr in K2Cr2O7 is +6 Cr+6 + 6e- → Cr This means 1 mol of Cr+6 needs 6 moles of electrons (6 faradays) 𝐴𝑟 𝑥 𝐼 𝑥 𝑡 Mass of Cr = 𝑖𝑜𝑛𝑠 𝑥 96500 = 52 𝑥 25 𝑥 7200 6 𝑥 96500 = 16.1658 gram Name: Description: ...
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