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Kimia Tingkatan 5 Bab 1 Tindak Balas Redoks
Tugasan 1
1. Hitung nombor pengoksidaan bagi unsur-unsur yang digariskan berikut: TP 2
Calculate the oxidation numbers for the following underlined elements: 1
Contoh / Example: Penyelesaian / Solution:
KMnO 4 KMnO 4 = 0
(+1) + Mn + 4(–2) = 0 BAB
Mn = +7
+ (b) Fe 2 O 3 2–
(a) NH 4 (c) CO 3
NH 4 = +1 Fe 2 O 3 = 0 CO 3 = –2
N + 4(+1) = +1 2Fe + 3(–2) = 0 C + 3(–2) = –2
N = –3 Fe = +3 C = +4
(d) HOCl (e) K 2 Cr 2 O 7 (f) Na 2 S 2 O 3
HOCl = 0 K 2 Cr 2 O 7 = 0 Na 2 S 2 O 3 = 0
(+1) + (–2) + Cl = 0 2(+1) + 2Cr + 7(–2) = 0 2(+1) + 2S + 3(–2) = 0
Cl = +1 Cr = +6 S = +2
2. Berdasarkan nombor pengoksidaan yang diberikan, namakan sebatian mengikut sistem penamaan IUPAC. TP 4
Based on the given oxidation numbers, name the compounds according to the IUPAC nomenclature.
Sebatian Nombor pengoksidaan Sistem penamaan IUPAC
Compound Oxidation number IUPAC nomenclature
Contoh / Example : Fe = +2
Fe 2 O 3 Ferum(II) oksida / Iron(II) oxide
(a) Cu 2 O Cu = +1 Kuprum(I) oksida / Copper(I) oxide
(b) PbI 2 Pb = +2 Plumbum(II) iodida / Lead(II) iodide
(c) K 2 Cr 2 O 7 Cr = +6 Kalium dikromat(VI) / Potassium dichromate(VI)
5. Seketul zink, Zn diletakkan ke dalam larutan kuprum(II) sulfat, CuSO 4 yang berwarna biru. Setelah 30 minit,
larutan biru bertukar menjadi tak berwarna.
A piece of zinc, Zn is placed into a blue copper(II) sulphate, CuSO 4 solution. After 30 minutes, the blue solution turns colourless.
(a) Nyatakan persamaan kimia seimbang bagi tindak balas redoks di atas. TP 2
State the balanced chemical equation for the above redox reaction.
Zn(p) + CuSO 4 (ak) → ZnSO 4 (ak) + Cu(p)
Zn(s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu(s)
(b) Nyatakan persamaan ion bagi tindak balas redoks di atas. TP 2
State the ionic equation for the above redox reaction.
2+
Zn + Cu → Zn + Cu
2+
(c) Tulis persamaan setengah yang menunjukkan tindak balas pengoksidaan dan penurunan. TP 2
Write the half-equations that show the oxidation and reduction reactions.
2+
• Pengoksidaan / Oxidation: Zn → Zn + 2e –
–
2+
• Penurunan / Reduction: Cu + 2e → Cu
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