Page 13 - Get Ready SPM 2022 Tg 5 - Matematik

Jawapan

Ujian 1 • Ferum mempunyai lebih Oxidising agent = manganate(VII)
daripada satu nombor ion / potassium manganate(VII)
Kertas 1 pengoksidaan manakala zink solution
hanya mempunyai satu nombor • Agen penurunan = ion bromida /
1. B 2. C 3. A 4. C 5. B pengoksidaan. larutan kalium bromida
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6. B 7. C 8. C 9. A 10. A Iron has more than one oxidation Reducing agent = bromide ion /
number while zinc has only one potassium bromide solution
Kertas 2 oxidation number.
(c) • Air bromin mengalami tindak Bahagian C
Bahagian A balas penurunan kerana nombor
pengoksidaan bromin berkurang 4. (a) • Terminal positif / Positive terminal
1. (a) Cu + 2e → Cu daripada 0 kepada -1. = Ag
2+
–
(b) 0 → +2 Bromine water undergoes • Terminal negatif / Negative
(c) Serbuk magnesium // Serbuk reduction reaction because terminal = Fe
aluminium bromine's oxidation number • Pengoksidaan / Oxidation:
Magnesium powder // Aluminium decreases from 0 to –1. Fe → Fe + 2e –
2+
powder • Ion iodida mengalami tindak • Penurunan / Reduction:
(d) Tiada perubahan. Argentum balas pengoksidaan kerana Ag + e → Ag
–
+
kurang elektropositif berbanding nombor pengoksidaannya (b) • Anod / Anode: Al
dengan kuprum. / Nilai E 0 meningkat daripada –1 kepada
argentum lebih positif berbanding 0. • Katod / Cathode: Cu
dengan kuprum. Iodide ion undergoes oxidation • Nilai voltan keseluruhan / Overall
No change. Silver is less reaction because iodine's oxidation voltage value
electropositive than copper. / The number increases from –1 to 0. = 0.34 – (‒1.66)
E 0 value of silver is more positive • Br 2 + 2I → 2Br + I 2 = 2.00 V
–
–
than copper. (d) • Ion bromida mengalami tindak (c) Prosedur / Procedure:
2. (a) 0.00 V balas pengoksidaan dengan 1. Bersihkan jalur logam X, Y dan
(b) W , Z , X , Y + melepaskan elektron untuk Z dengan kertas pasir.
2+
3+
2+
(c) (i) W membentuk bromin. Clean metal strips X, Y and Z
(ii) E sel = E katod – E anod Bromide ion undergoes oxidation with sandpaper.
0
0
0
3
E cell = E cathode – E anode reaction by releasing electrons 2. Tuang 5 cm larutan X nitrat
0
0
0
forming bromine. 1.0 mol dm ke dalam tiga
–3
–
= 0.34 – (‒2.38) • 2Br → Br 2 + 2e – tabung uji yang berasingan.
–3
3
= 2.72 V • Ion manganat(VII) mengalami Pour 5 cm of 1.0 mol dm
balas
penurunan
tindak
(iii) 0.46 V dengan menerima elektron X nitrate solution into three
Y bertindak sebagai katod dalam keadaan berasid untuk separate test tubes.
kerana kurang elektropositif membentuk ion mangan(II) dan 3. Masukkan setiap logam ke
berbanding dengan X. dalam tabung uji berisi larutan
Y acts as a cathode because it air. ion undergoes X nitrat.
Manganate(VII)
is less electropositive than X. Put each metal into test tubes
(d) reduction reaction by accepting containing X nitrate solution.
electrons in acidic condition to
form mangan(II) ion and water. 4. Biarkan selama 5 minit.
V • MnO 4 + 8H + 5e → Mn +
‒
–
Leave on for 5 minutes.
+
2+
e – 4H 2 O 5. Catatkan pemerhatian dalam
• Elektron bergerak dari elektrod setiap tindak balas dalam
X-Terminal Z di dalam larutan kalium bromida jadual.
positif ke elektrod di dalam larutan Record the observation in each
X-Positive Larutan kalium manganat(VII) melalui reaction in a table.
terminal Z nitrat litar luar. 6. Ulang eksperimen dengan
Pasu berliang Z nitrate Electrons flow from the electrode menggantikan larutan X nitrat
Porous pot solution in potassium bromide solution dengan larutan Y nitrat dan
Larutan X to the electrode in potassium larutan Z nitrat.
sulfat manganate(VII) solution through Repeat the experiment by
X sulphate an external circuit.
solution • Agen pengoksidaan = ion replacing X nitrate solution with
manganat(VII) / larutan kalium Y nitrate solution and Z nitrate
manganat(VII) solution.
Bahagian B
3. (a) • Mg + ZnSO 4 → MgSO 4 + Zn Logam Larutan X nitrat Larutan Y nitrat Larutan Z nitrat
• Pengoksidaan / Oxidation: Mg Metal X nitrate solution Y nitrate solution Z nitrate solution
• Penurunan / Reduction: Zn / Tiada tindak balas Tiada tindak balas Tiada tindak balas
2+
ZnSO 4 X No reaction No reaction No reaction
(b) • FeSO 4 = Ferum(II) sulfat / Iron(II) Logam X disesarkan Tiada tindak balas Tiada tindak balas
sulphate Y
• ZnSO 4 = Zink sulfat / Zinc X metal is displaced No reaction No reaction
sulphate Z Logam X disesarkan Logam Y disesarkan Tiada tindak balas
X metal is displaced Y metal is displaced No reaction
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