Transformations Rapides Transformations Lentes
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ا ــــــت
ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺴﺮﻳﻌﺔ ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺒﻄﻴﺌﺔ Iﺗﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ -ﺍﺧﺘﺰﺍﻝ: (1ﺍﻹﺑﺮﺍﺯ ﺍﻟﺘﺠﺮﻳﱯ ﳌﻔﻬﻮﻡ ﺍﻷﻛﺴﺪﺓ -ﺍﺧﺘﺰﺍﻝ: ﺃ(ﲡﺮﺑﺔ:
ﻧﻐﻤﺮ ﺻﻔﻴﺤﺔ ﻣﻦ ﺍﻟﻨﺤﺎﺱ ﰲ ﻛﺄﺱ ﺑﻪ ﳏﻠﻮﻝ ﻣﺎﺋﻲ ﻟﻨﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ) ( Ag + + NO3 − aq
aq
ﻧﻼﺣﻆ ﺗﻮﺿﻊ ﻃﺒﻘﺔ ﻣﻦ ﺍﻟﻔﻀﺔ Agﻋﻠﻰ ﺍﳉﺰﺀ ﺍﳌﻐﻤﻮﺭ ﻭﺗﻠﻮﻥ ﺍﶈﻠﻮﻝ ﺑﺎﻟﻠﻮﻥ ﺍﻷﺯﻕ ﻧﺘﻴﺠﺔ ﺗﻜﻮﻥ ﺍﻷﻳﻮﻧﺎﺕ . Cu 2+ ) ( aq
) (s
ﺏ(ﺗﻌﻠﻴﻞ: ⊗ ﺧﻼﻝ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺗﺄﻛﺴﺪﺕ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ ﺇﱃ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻨﺤﺎﺱ ﺣﺴﺐ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ : 2e −
+
Cu 2+ ) ( aq
ﺃﻛﺴﺪﺓ
Cu ) (s
ﻭﻫﻜﺬﺍ ﻟﻌﺒﺖ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ ﺩﻭﺭ ﺍﳌﺨﺘﺰﻝ)ﻷﺎ ﻓﻘﺪﺕ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ( ﻓﺘﺤﻮﻟﺖ ﺇﱃ ﺍﳌﺆﻛﺴﺪ ﺍﳌﺮﺍﻓﻖ Cu 2+ﻭﻫﺬﺍﻥ ﺍﻟﻨﻮﻋﺎﻥ ) ( aq
ﻳﻨﺘﻤﻴﺎﻥ ﺇﱃ ﺍﳌﺰﺩﻭﺝ . Cu 2+ / Cu )(s
aq
⊗ ﺑﻴﻨﻤﺎ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻔﻀﺔ ﺍﺧﺘﺰﻟﺖ ﺇﱃ Agﺣﺴﺐ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ : ) (s
Ag
ﺍﺧﺘﺰﺍﻝ
e−
) (s
+
+
Ag
) (aq
ﲝﻴﺚ ﻟﻌﺒﺖ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻔﻀﺔ Ag +ﺩﻭﺭ ﺍﳌﺆﻛﺴﺪ)ﻷﺎ ﺍﻛﺘﺴﺒﺖ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ( ﻓﺘﺤﻮﻟﺖ ﺇﱃ ﺍﳌﺨﺘﺰﻝ ﺍﳌﺮﺍﻓﻖ Agﻭﻫﺬﺍﻥ ﺍﻟﻨﻮﻋﺎﻥ ﻳﻨﺘﻤﻴﺎﻥ ) (s
) (aq
ﺇﱃ ﺍﳌﺰﺩﻭﺟﺔ . Ag + / Ag ) (s
) (aq
1
⊗ ﻭﳝﻜﻦ ﺍﳊﺼﻮﻝ ﻋﻠﻰ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﲜﻤﻊ ﻧﺼﻔﻲ ﻣﻌﺎﺩﻟﱵ ﺍﻷﻛﺴﺪﺓ –ﺍﺧﺘﺰﺍﻝ : 2e −
Cu 2+
+
ﺍﻛﺴﺪﺓ
) ( aq
Cu ) (s
ﺍﺧﺘﺰﺍﻝ e −
Ag ) × 2
( Ag + + aq
) (s
+ 2 Ag +
Cu 2+
+ 2 Ag
) ( aq
) (s
) (aq
Cu ) (s
ﺩ( ﺍﺳﺘﻨﺘﺎﺝ: ﺍﳌﺆﻛﺴﺪﻫﻮ ﻛﻞ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻗﺎﺩﺭ ﻋﻠﻰ ﺍﻛﺘﺴﺎﺏ ﺍﻟﻜﺘﺮﻭﻥ ﺃﻭ ﺃﻛﺜﺮ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ،ﻭﻳﺴﻤﻰ ﺍﻟﻨﻮﻉ ﺍﻟﻨﺎﺗﺞ ﳐﺘﺰﻻ ﻣﺮﺍﻓﻘﺎ . ﺍﳌﺨﺘﺰﻝ ﻫﻮ ﻛﻞ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻗﺎﺩﺭ ﻋﻠﻰ ﻣﻨﺢ ﺍﻟﻜﺘﺮﻭﻥ ﺃﻭ ﺃﻛﺜﺮ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ،ﻭﻳﺴﻤﻰ ﺍﻟﻨﻮﻉ ﺍﻟﻨﺎﺗﺞ ﻣﺆﻛﺴﺪﺍ ﻣﺮﺍﻓﻘﺎ . ﺍﻷﻛﺴﺪﺓ ﻫﻲ ﻓﻘﺪﺍﻥ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ﻣﻦ ﻃﺮﻑ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻭﺍﻟﻨﻮﻉ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺬﻱ ﺗﻄﺮﺃ ﻋﻠﻴﻪ ﺍﻷﻛﺴﺪﺓ ﻳﺴﻤﻰ ﳐﺘﺰﻻ. ﻭﺍﻹﺧﺘﺰﺍﻝ ﻫﻮ ﺍﻛﺘﺴﺎﺏ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ﻣﻦ ﻃﺮﻑ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻭﺍﻟﻨﻮﻉ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺬﻱ ﻳﻄﺮﺃ ﻋﻠﻴﻪ ﺍﻹﺧﺘﺰﺍﻝ ﻳﺴﻤﻰ ﻣﺆﻛﺴﺪﺍ. ﺑﺼﻔﺔ ﻋﺎﻣﺔ ﻳﺘﻢ ﺗﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﲔ ﻣﺰﺩﻭﺟﺘﲔ ﳐﺘﺰﻝ/ﻣﺆﻛﺴﺪ ox1 / réd1ﻭ ox 2 / réd 2ﲝﻴﺚ ﳛﺪﺙ ﺍﻧﺘﻘﺎﻝ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ( ox1 + n1e − → réd1 ) × n2
ﻣﻦ ﳐﺘﺰﻝ ﺇﺣﺪﺍ ﳘﺎ ﺇﱃﻣﺆﻛﺴﺪ ﺍﻻﺧﺮﻯ
(réd 2 → ox 2 + n 2 e − ) × n1 ←
n 2 ox1 + n1 réd 2 → n 2 réd 1 + n1ox 2
ل :ﻋﻨﺩﻤﺎ ﻨﻐﻤﺭ ﺼﻔﻴﺤﺔ ﻤﻥ ﺍﻷﻝﻤﻨﻴﻭﻡ ﻓﻲ ﻤﺤﻠﻭل ﻤﺎﺌﻲ ﻝﻨﺘﺭﺍﺕ ﺍﻝﻔﻀﺔ ﻨﺤﺼل ﻋﻠﻰ ﺘﻭﻀﻊ ﺍﻝﻔﻀﺔ ﻋﻠﻰ ﺍﻝﺼﻔﻴﺤﺔ ﻭﻨﺒﺭﺯ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻝﺘﻔﺎﻋل ﺘﻜﻭﻥ ﺃﻴﻭﻨﺎﺕ ﺍﻷﻝﻤﻨﻴﻭﻡ .ﺇﺫﻥ ﺍﻝﺘﻔﺎﻋل ﺍﻝﺫﻱ ﺘﻡ ﺒﻴﻥ ﺍﻝﻤﺯﺩﻭﺠﺘﻴﻥ Al 3+ / Al )(s
) ( aq
و Ag + / Agﻫﻮ ﻛﻤﺎ ﻳﻠﻲ: )(s
) ( aq −
Ag ) × 3
+
( Ag + e
Al 3+ + 3e − 3 Ag + Al 3+
Al Al + 3 Ag +
(2ﺇﺛﺒﺎﺕ ﻣﻌﺎﺩﻟﺔ ﺗﻔﺎﻋﻞ ﺍﻟﻸﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ: ﺃ(
ﲡﺮﺑﺔ:ﻧﺼﺐ ﰲ ﻛﺄﺱ ﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﻛﱪﻳﺘﺎﺕ ﺍﳊﺪﻳﺪ
II
) . ( Fe 2+ + SO4 2−
ﰒ ﻧﻀﻴﻒ ﺇﻟﻴﻪ ﻗﻄﺮﺓ ،ﻗﻄﺮﺓ ،ﳏﻠﻮﻝ ﺑﺮﻣﻐﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ) ( K + + MnO4 −ﺍﶈﻤﺾ ﲝﻤﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ.
ﻧﻼﺣﻆ ﺃﻥ ﳏﻠﻮﻝ ﺑﺮﻣﻐﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ﻳﻔﻘﺪ ﻟﻮﻧﻪ ﺍﻟﺒﻨﻔﺴﺠﻲ ﻧﺘﻴﺠﺔ ﺗﻜﻮﻥ ﺃﻳﻮﻧﺎﺕ ﺍﳌﻨﻐﻨﻴﺰ Mn 2+ﺍﻟﻌﺪﳝﺔ ﺍﻟﻠﻮﻥ ﰲ ﺍﶈﺎﻟﻴﻞ ) ( aq
ﺍﳌﺎﺋﻴﺔ. ﻭ ﻧﱪﺯ ﺑﺎﺳﺘﻌﻤﺎﻝ ﳏﻠﻮﻝ ﺍﻟﺼﻮﺩﺍ ﻭﺟﻮﺩ ﺍﻷﻳﻮﻧﺎﺕ Fe 3+ﰲ ﺍﶈﻠﻮﻝ ﺍﶈﺼﻞ ﻋﻠﻴﻪ)ﺍﻟﱵ ﺗﻌﻄﻲ ﻟﻠﻤﺤﻠﻮﻝ ﻟﻮﻥ ﺍﻟﺼﺪﺃ(. ) ( aq
ﺧﻼﻝ ﻫﺬﻩ ﺍﻟﺘﺠﺮﺑﺔ ﰎ ﺗﻔﺎﻋﻞ ﺑﲔ ﺃﻳﻮﻧﺎﺕ ﺍﻟﱪﻣﻨﻐﻨﺎﺕ MnO4 −ﻭﺃﻳﻮﻧﺎﺕ
ﺍﳊﺪﻳﺪII
) ( aq
ﺍﳊﺪﻳﺪ III
. Fe 3+ ) ( aq
ﺏ( ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ: ﻟﻨﻜﺘﺐ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﻭﺟﺔ
: Fe 3+ / Fe 2+ ) ( aq
) ( aq
2
2+
Feﻹﻋﻄﺎﺀ ﺃﻳﻮﻧﺎﺕ ﺍﳌﻨﻐﻨﻴﺰ
) ( aq
2+
Mnﻭﺃﻳﻮﻧﺎﺕ
) ( aq
Fe 3+ + e −
Fe 2+
) ( aq
−
/ Mn 2+
ﻭﻟﻨﻜﺘﺐ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﻭﺟﺔ
) ( aq
) ( aq
:
MnO4 ) ( aq
ﺑﺼﻔﺔ ﻋﺎﻣﺔ ﳌﻮﺍﺯﻧﺔ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ ﺃﻭ ﺍﻹﺧﺘﺰﺍﻝ ﻧﺘﺘﺒﻊ ﺍﳋﻄﻮﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ : ﻧﻮﺍﺯﻥ ﺍﻟﻌﻨﺼﺮ ﺍﻟﺬﻱ ﺗﻌﺮﺽ ﻟﺘﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ ﺃﻭ ﺍﻹﺧﺘﺰﺍﻝ. ﻧﻮﺍﺯﻥ ﻋﻨﺼﺮ ﺍﻷﻭﻛﺴﺠﲔ ﺑﺈﺿﺎﻓﺔ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ . H 2 O ﻧﻮﺍﺯﻥ ﻋﻨﺼﺮ ﺍﳍﻴﺪﺭﻭﺟﲔ ﺑﺈﺿﺎﻓﺔ ﺍﻟﱪﻭﺗﻮﻧﺎﺕ . H + ﳓﻘﻖ ﻣﻮﺍﺯﻧﺔ ﺍﳌﻌﺎﺩﻟﺔ ﻣﻦ ﺣﻴﺚ ﺍﻟﺸﺤﻨﺎﺕ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺑﺈﺿﺎﻓﺔ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ﺟﻬﺔ ﺍﳌﺆﻛﺴﺪ.ﻓﻨﺤﺼﻞ ﻋﻠﻰ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: + 4 H 2O )(l
−
MnO4 + 8 H + + 5e −
Mn 2+ +
) ( aq
) ( aq
) ( aq
ﻭﳓﺼﻞ ﻋﻠﻰ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﺈﺿﺎﻓﺔ ﻧﺼﻔﻲ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ .
ﺗﻄﺒﻴﻖ:ﺍﻛﺘﺐ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ -ﺍﺧﺘﺰﺍﻝ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ / Cr 3+
ﺍﳌﺮﺣﻠﺔ ﺍﻷﻭﱃ:
2 Cr 3+
ﺍﳌﺮﺣﻠﺔ ﺍﻟﺜﺎﻧﻴﺔ:
7H 2O
2−
Cr2 O7
2−
Cr2 O7
2 Cr 3+
+
+ 14 H + + 6e −
2−
Cr2 O7
IIﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺴﺮﻳﻌﺔ ﻭﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺒﻄﻴﺌﺔ: (1ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺴﺮﻳﻌﺔ: ﺃ( ﺗﻌﺮﻳﻒ: ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺴﺮﻳﻌﺔ ﻫﻲ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﱵ ﲢﺪﺙ ﰲ ﻣﺪﺓ ﺯﻣﻨﻴﺔ ﻭﺟﻴﺰﺓ ﲝﻴﺚ ﻻ ﳝﻜﻦ ﺗﺘﺒﻊ ﺗﻄﻮﺭﻫﺎ ﺑﺎﻟﻌﲔ ﺍﺮﺩﺓ ﺃﻭ ﺑﺄﺩﻭﺍﺕ ﺍﻟﻘﻴﺎﺱ .
ﺏ( ﺃﻣﺜﻠﺔ: ∗ ﻣﺜﺎﻝ:1ﺗﺮﺳﻴﺐ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﺤﺎﺱ + 2 OH −
Cu (OH ) 2
) ( aq
) ( aq
C u 2+ ) ( aq
∗ ﻣﺜﺎﻝ :2ﺗﺮﺳﻴﺐ ﻛﻠﻮﺭﻭﺭ ﺍﻟﻔﻀﺔ + Cl −
) ( AgCl
) (aq
)(s
Ag + ) (aq
∗ ﻣﺜﺎﻝ:3 ﺍﻟﺘﻔﺎﻋﻞ ﺑﲔ ﺍﳌﺰﺩﻭﺟﺘﲔ / S 2 O3 2− : ) ( aq
2−
S 4 O6
ﻭ
) ( aq
. I2 / I −
) (aq
) ( aq
ﻧﺼﺐ ﰲ ﻛﺄﺱ ﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﺛﻴﻮ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ) (2 Na + + S 2 O3 2−ﻧﻀﻴﻒ ﺇﻟﻴﻪ ﺗﺪﺭﳚﻴﺎ ﳏﻠﻮﻝ ﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ . I 2 ) ( aq
3
) ( aq
) ( aq
ﻳﺪﻝ ﺍﻹﺧﺘﻔﺎﺀ ﺍﻟﻠﺤﻈﻲ ﻟﻠﻮﻥ ﺍﻟﺒﲏ ﻟﺜﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ ﻋﻠﻰ ﺃﻥ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺳﺮﻳﻊ. ﺧﻼﻝ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺗﺄﻛﺴﺪﺕ ﺃﻳﻮﻧﺎﺕ ﺛﻴﻮ ﻛﱪﻳﺘﺎﺕ S 2 O3 2−ﺇﱃ ﺃﻳﻮﻧﺎﺕ ﺛﻴﻮﻧﺎﺕ S 4 O6 2−ﻭﻓﻖ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: ) ( aq 2−
+ 2e −
) ( aq 2−
ﺃﻛﺴﺪﺓ
S 4 O6 ) ( aq
2 S 2 O3 ) ( aq
ﺑﻴﻨﻤﺎ ﺍﺧﺘﺰﻟﺖ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻴﻮﺩ I 2ﺇﱃ I −ﻭﻓﻖ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: ) ( aq ) ( aq
−
−
ﺍﺧﺘﺰﺍﻝ
2 I
) ( aq
+ 2e
I2 ) ( aq
ﻭﳓﺼﻞ ﻋﻠﻰ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﺈﺿﺎﻓﺔ ﻧﺼﻔﻲ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ ﻃﺮﻓﺎ ﺑﻄﺮﻑ. 2 I−
) ( aq
2−
+
2 S 2 O3 2− + I 2
S 4 O6 ) ( aq
) ( aq
) ( aq
(2ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺒﻄﻴﺌﺔ: ﺃ( ﺗﻌﺮﻳﻒ: ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺒﻄﻴﺌﺔ ﻫﻲ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﱵ ﺗﺴﺘﻐﺮﻕ ﻣﺪﺓ ﺯﻣﻨﻴﺔ ﻧﺴﺘﻄﻴﻊ ﺧﻼﳍﺎ ﺗﺘﺒﻌﻪ ﺑﺎﻟﻌﲔ ﺍﺮﺩﺓ ﺃﻭ ﺑﺄﺩﻭﺍﺕ ﺍﻟﻘﻴﺎﺱ .
ﺏ(ﻣﺜﺎﻝ:1 ﳕﺰﺝ ﰲ ﻛﺄﺱ ﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﻳﻮﺩﻭﺭ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ) ( K + + I −ﻭﺍﳌﺎﺀ ﺍﻷﻛﺴﻴﺠﻴﲏ H 2 O2ﺍﶈﻤﺾ ﺑﻘﻄﺮﺍﺕ ﻣﻦ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ﺍﳌﺮﻛﺰ . ﻣﺒﺎﺷﺮﺓ ﺑﻌﺪ ﻣﺰﺝ ﺍﶈﻠﻮﻟﲔ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ﺑﲔ I −ﻭ H 2 O2ﻓﻴﺘﻜﻮﻥ ﺗﺪﺭﳚﻴﺎ ﺍﳌﺎﺀ ﻭﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ I 2ﺫﻱ ﺍﻟﻠﻮﻥ ﺍﻟﺒﲏ. H 2 O 2 + 2 H + + 2e − → 2 H 2 O
ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ : H 2 O2 / H 2 O ) (l
ﻭﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ
) ( aq
) ( aq
) (l
: I2 / I −
) ( aq
2 I − → I 2 + 2e −
) ( aq ) ( aq
) ( aq
) ( aq
ﻭﳓﺼﻞ ﻋﻠﻰ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﺈﺿﺎﻓﺔ ﻧﺼﻔﻲ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ ﻃﺮﻓﺎ ﺑﻄﺮﻑ. H 2 O2 + 2 I − + 2 H + → I 2 + 2 H 2 O ) (l
) ( aq
) ( aq
ﳝﻜﻦ ﺗﺘﺒﻊ ﺗﻄﻮﺭ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﲟﻌﺎﻳﺮﺓ ﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ I 2ﺍﳌﻜﻮﻥ ﺑﻮﺍﺳﻄﺔ ﺃﻳﻮﻧﺎﺕ ﺛﻴﻮﻛﱪﻳﺘﺎﺕ
) ( aq
2−
) ( aq
. S 2 O3 ) ( aq
) ( aq
ﺝ( ﻣﺜﺎﻝ:2 ﳕﺰﺝ ﰲ ﻛﺄﺱ 10mlﻣﻦ ﳏﻠﻮﻝ ﲪﺾ ﺍﻟﻜﻠﻮﺭﻳﺪﺭﻳﻚ ) ( H + + Cl −ﺍﳌﺮﻛﺰ) (1mol / lﻭﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﺛﻴﻮ ﻛﱪﻳﺘﺎﺕ ) ( aq
) ( aq
ﺍﻟﺼﻮﺩﻳﻮﻡ ) (2 Na + + S 2 O3 2−ﺫﻱ ﺍﻟﺘﺮﻛﻴﺰ . 10 −1 mol / l ) ( aq
) ( aq
ﻧﻼﺣﻆ ﺃﻥ ﳏﺘﻮﻯ ﺍﻟﻜﺄﺱ ﻳﺄﺧﺬ ﺑﻌﺪ ﳊﻈﺎﺕ ﻟﻮﻧﺎ ﺃﺻﻔﺮﺍﻧﺘﻴﺠﺔ ﺗﻜﻮﻥ ﺍﻟﻜﱪﻳﺖ ﻭﻫﺬﺍ ﺍﻟﺘﺤﻮﻝ ﺑﻄﻲﺀ ﻷﻧﻪ ﻳﺴﺘﻐﺮﻕ ﻣﺪﺓ ﺗﺴﻤﺢ ﺑﺘﺘﺒﻌﻪ ،ﺧﺎﺻﺔ ﻋﻨﺪﻣﺎ ﻧﺴﻠﻂ ﻋﻠﻴﻪ ﺣﺰﻣﺔ ﻣﻦ ﺍﻟﻀﻮﺀ ﺍﻷﺑﻴﺾ ﻭﺩﻗﺎﺋﻖ ﺍﻟﻜﱪﻳﺖ ﺍﻟﻌﺎﻟﻘﺔ ﰲ ﺍﶈﻠﻮﻝ ﺗﺸﺘﺖ ﺍﻟﻀﻮﺀ ﻭﺧﺎﺻﺔ ﺍﳌﻮﺟﺎﺕ ﺍﳌﻮﺍﻓﻘﺔ ﻟﻠﻮﻥ ﺍﻷﺯﺭﻕ ﻓﻴﺄﺧﺬ ﺍﳋﻠﻴﻂ ﰲ ﺍﻟﺒﺪﺍﻳﺔ ﻟﻮﻧﺎ ﺃﺯﺭﻗﺎ ﰒ ﻳﺘﺤﻮﻝ ﺇﱃ ﺍﻷﺻﻔﺮ.
4
ﻳﺘﻤﻴﺰ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﺯﺩﻭﺍﺟﻴﺔ ﲢﻮﻝ ﺍﻷﻳﻮﻧﺎﺕ S 2 O3 2−ﰲ ﻭﺳﻂ ﲪﻀﻲ ﻋﻠﻰ ﺇﱃ Sﰒ ﺇﱃ ) . ( SO ) (S
) ( aq
: S 2 O3 2− / S
ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ + 3 H 2O
S
) (l
) (S
) ( aq
+ 4e −
ﺍﺧﺘﺰﺍﻝ
2
) (S
2−
ﻭﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ / S 2 O3 : 4e −
2H + + ) ( aq
) ( aq
) ( aq
SO
ﺃﻛﺴﺪﺓ
2 SO
+ 6H +
2−
S 2 O3
2 ) ( aq
) ( aq
+
2 ) ( aq
+
H 2O
2 ) ( aq
2−
) (l
S 2 O3 ) ( aq
ﻭﺣﺼﻴﻠﺔ ﺍﻟﺘﻔﺎﻋﻞ ﻫﻲ: + SO
+ H 2O
2 ) ( aq
) (l
2H +
S
) ( aq
) (S
2−
+
S 2 O3 ) ( aq
ﺩ( ﺃﻣﺜﻠﺔ ﺍﺧﺮﻯ ﻟﺘﺤﻮﻻﺕ ﺑﻄﻴﺌﺔ : ﺗﻔﺎﻋﻞ ﺍﻷﺳﺘﺮﺓ.-ﺗﻜﻮﻥ ﺍﻟﺼﺪﺃ ﺍﻟﺬﻱ ﻳﻨﺘﺞ ﻋﻨﻪ ﺗﺂﻛﻞ ﺍﳊﺪﻳﺪ.
ﻣﻠﺤﻮﻇﺔ:1
ﺗﻌﺘﱪ ﺑﻌﺾ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺟﺪ ﺑﻄﻴﺌﺔ ﲝﻴﺚ ﺗﻈﻬﺮ ﻭﻛﺄﺎ ﻻﲢﺪﺙ،ﻧﻘﻮﻝ ﺇﺎ ﺣﺮﻛﻴﺎ ﻣﺘﻮﻗﻔﺔ.
ﻣﺜﺎﻝ :ﲢﻠﻞ ﺍﳌﺎﺀ ﺍﻷﻛﺴﻴﺠﻴﲏ
+ O2 )(g
H 2 O2
H 2O ) (l
) ( aq
ﻣﻠﺤﻮﻇﺔ :2ﻹﺑﺮﺍﺯ ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺒﻄﻴﺌﺔ ﻧﺴﺘﻌﻤﻞ ﺑﻌﺾ ﺍﻟﺘﻘﻨﻴﺎﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ،ﻣﺜﻞ:ﺍﺳﺘﻌﻤﺎﻝ ﺍﳌﺎﻧﻮﻣﻴﺘﺮ ﺃﻭﻣﻘﻴﺎﺱ ﺍﳌﻮﺍﺻﻠﺔﺃﻭﺍﺳﺘﻌﻤﺎﻝ ﺟﻬﺎﺯ - pHﻣﻴﺘﺮ.
(IIIﺍﻹﺑﺮﺍﺯ ﺍﻟﺘﺠﺮﻳﱯ ﻟﻠﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ: (1ﺗﻌﺮﻳﻒ: ﻧﺴﻤﻲ ﻋﺎﻣﻼ ﺣﺮﻛﻴﺎ ﻛﻞ ﻣﻘﺪﺍﺭ ﻣﺆﺛﺮ ﻋﻠﻰﺳﺮﻋﺔ ﲢﻮﻝ ﻛﻴﻤﻴﺎﺋﻲ .
(2ﺃﻣﺜﻠﺔ ﻟﺒﻌﺾ ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ :
ﺃ(ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ : ﻧﻨﺠﺰ ﺍﻟﺘﺠﺮﺑﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻹﺑﺮﺍﺯ ﺗﺄﺛﲑ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﻋﻠﻰ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ . ﻧﺼﺐ ﰲ ﺃﻧﺒﻮﰊ ﺇﺧﺘﺒﺎﺭ )(1ﻭ) 10ml (2ﻣﻦ ﳏﻠﻮﻝ ﲪﺾ ﺍﻷﻭﻛﺴﺎﻟﻴﻚ H 2 C 2 O4ﺫﻱ ﺗﺮﻛﻴﺰ . 0,5mol / l ) ( aq
ﻧﺘﺮﻙ ﺍﻟﻜﺄﺱ ) (1ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻋﺎﺩﻳﺔ ﻭﻧﺮﻓﻊ ﺩﺭﺟﺔ ﺍﻟﺜﺎﱐ ﺇﱃ . 40°c ﻧﻀﻴﻒ ﰲ ﻧﻔﺲ ﺍﻟﻠﺤﻈﺔ ﺇﱃ ﻛﻞ ﻣﻨﻬﻤﺎ 5mlﺑﺮﻣﻐﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ) . ( K + + MnO4 − ﻧﻼﺣﻆ ﺍﺧﺘﻔﺎﺀ ﺍﻟﻠﻮﻥ ﺍﻟﺒﻨﻔﺴﺠﻲ ﺍﳌﻤﻴﺰ ﻟﻸﻳﻮﻧﺎﺕ MnO4 −ﺃﺳﺮﻉ ﻗﻲ ﺍﻟﻜﺄﺱ 2 ﺍﻟﺬﻱ ﰎ ﺗﺴﺨﻴﻨﻪ )ﺃﻱ ﺭﻓﻊ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺗﻪ(. ﻧﺴﺘﻨﺘﺞ ﺃﻧﻪ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﻣﺮﺗﻔﻌﺔ ﻛﻠﻤﺎ ﺍﺯﺩﺍﺩﺕ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ.
ﺏ( ﺗﺎﺛﲑ ﺗﺮﺍﻛﻴﺰ ﺍﳌﺘﻔﺎﻋﻼﺕ : 5
1
2
ﺗﺘﻔﺎﻋﻞ ﺍﻳﻮﻧﺎﺕ ﺍﻟﻴﻮﺩﻭﺭ I −ﻣﻊ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ ﺍﻷﻛﺴﻴﺠﻴﲏ H 2 O2ﰲ ﻭﺳﻂ ﲪﻀﻲ ﺣﺴﺐ ﺍﳌﻌﺎﺩﻟﺔ : H 2 O2 + 2 I − + 2 H + → I 2 + 2 H 2 O ) (l
) ( aq
) ( aq
) ( aq
) ( aq
ﻧﺼﺐ ﰲ ﻛﺄﺳﲔ ﻧﻔﺲ ﺍﳊﺠﻢ H 2 O2ﺍﶈﻤﺾ ﺑﻘﻄﺮﺍﺕ ﻣﻦ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ﻭﻧﻀﻴﻒ ﻟﻜﻞ ﻣﻨﻬﻤﺎ ﻧﻔﺲ ﺍﳊﺠﻢ ﻣﻦ ﳏﻠﻮﻝ ) ( K + + I −ﻟﻜﻦ ﺗﺮﻛﻴﺰ ﻫﺬﺍ ﺍﻻﺧﲑ ﰲ ﺍﻟﻜﺎﺱ ﺍﻟﺜﺎﱐ ﺃﻛﱪ ﻣﻨﻪ ﰲ ﺍﻟﻜﺄﺱ ﺍﻷﻭﻝ . ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻳﻜﻮﻥ ﺃﺳﺮﻉ ﻛﻠﻤﺎ ﻛﺎﻥ ﺗﺮﻛﻴﺰﺍﺣﺪ ﺍﳌﺘﻔﺎﻋﻼﺕ ﺍﻛﱪ. ﻛﻤﺎ ﻧﺸﲑ ﺍﱃ ﺍﻥ ﻫﻨﺎﻙ ﻋﻮﺍﻣﻞ ﺍﺧﺮﻯ ﻣﺜﻞ ﺍﳊﻔﺎﺯ ﻭﺍﳌﺬﻳﺐ .
(3ﺗﻔﺎﻋﻼﻥ ﺧﺎﺻﺎﻥ: ﺃ( ﻣﺘﻔﺎﻋﻞ ﻓﻬﻠﲔ :ﺗﻔﺎﻋﻞ ﺧﺎﺹ ﺑﻄﻲﺀ ﻳﺴﺘﻌﻤﻞ ﻛﺮﺍﺋﺰ ﻟﺘﻤﻴﻴﺰ ﺍﻷﻟﺪﻫﻴﺪﺍﺕ . ﳛﺘﻮﻱ ﳏﻠﻮﻝ ﺍﻟﻔﻬﻠﲔ ﻋﻠﻰ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻨﺤﺎﺱ IIﻣﻌﻘﺪ ﰲ ﻭﺳﻂ ﻗﺎﻋﺪﻱ ﺗﻌﻄﻲ ﻟﻠﻤﺤﻠﻮﻝ ﻟﻮﻧﺎ ﺃﺯﺭﻗﺎ ﺩﺍﻛﻨﺎ،ﻭﻟﻠﺘﺒﺴﻴﻂ ﻧﺮﻣﺰ ﳍﺬﻩ ﺍﻷﻳﻮﻧﺎﺕ ﺏ Cu 2+ ( aq ) :ﻭﻫﻲ ﺍﻟﱵ ﺗﻜﻮﻥ ﻣﺆﻛﺴﺪ ﺍﳌﺰﺩﻭﺟﺔ. Cu 2+ / Cu 2 O : ) ( aq
) ( aq
ﻧﺼﺐ ﰲ ﺃﻧﺒﻮﺏ ﺍﺧﺘﺒﺎﺭ ﻗﻠﻴﻼ ﻣﻦ ﺍﻟﻔﻬﻠﲔ ﻭﻧﻀﻴﻒ ﺇﻟﻴﻬﺎ ﻗﻠﻴﻼ ﻣﻦ ﺍﻹﻳﺜﺎﻧﺎﻝ CH 3CHOﻭﻫﻮ ﺃﻟﺪﻫﻴﺪ. ﺍﻟﺘﻔﺎﻋﻞ ﺍﳊﺎﺻﻞ ﺟﺪ ﺑﻄﻲﺀ ﻟﻜﻦ ﻋﻨﺪ ﺗﺴﺨﻴﻨﻪ ﺑﺒﻂﺀ ﻳﺘﻜﻮﻥ ﺍﻭﻛﺴﻴﺪ ﺍﻟﻨﺤﺎﺱ Cu 2 Oﻋﻠﻰ ﺷﻜﻞ ﺭﺍﺳﺐ ﺫﻱ ﻟﻮﻥ ﺃﲪﺮ ﺁﺟﻮﰊ . ﺇﺫﻥ ﰎ ﺗﻔﺎﻋﻞ ﺑﲔ ﺍﳌﺰﺩﻭﺟﺘﲔ :
CH3COO− / CH3CHO
ﻭ. Cu 2+ / Cu 2 O : ) ( aq
) ( aq
CH 3CHO + 3 OH − → CH 3 COO − + 2 H 2 O + 2e − ) (l
) ( aq
) ( aq −
) ( aq 2+
−
2 Cu + 2 OH + 2e → Cu 2 O + H 2 O ) (l
) ( aq
)(s −
→ CH 3COO + 3 H 2 O + Cu 2 O ) ( aq
) (l
2+
) ( aq
) ( aq
−
CH 3CHO + 5 OH + 2 Cu
) ( aq
ﺏ( ﻣﺘﻔﺎﻋﻞ ﺗﻮﻟﻴﻨﺲ
) ( aq
)ﺍﳌﺴﻤﻰ ﻛﺬﻟﻚ:ﻧﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ﺍﻷﻣﻮﻧﻴﺎﻛﻲ ( Ag (NH 3 ) 2 +ﺗﻔﺎﻋﻞ ﺁﺧﺮ
ﺧﺎﺹ ﺑﻄﻲﺀ ﻳﺴﺘﻌﻤﻞ ﻛﺮﺍﺋﺰ ﻟﺘﻤﻴﻴﺰ ﺍﻷﻟﺪﻫﻴﺪﺍﺕ . ﻳﺘﻔﺎﻋﻞ ﻫﺬﺍ ﺍﻟﺮﺍﺋﺰ ﻣﻊ ﺍﻹﺛﺎﻧﺎﻝ ﰲ ﻭﺳﻂ ﻗﺎﻋﺪﻱ ﻓﻴﺘﻜﻮﻥ ﺭﺍﺳﺐ ﻣﻦ ﺍﻟﻔﻀﺔ ﺍﻟﺬﻱ ﻳﺴﺘﻌﻤﻞ ﻟﺘﻔﻀﻴﺾ ﺍﳌﺮﺍﻳﺎ . ﺍﳌﺰﺩﻭﺟﺘﺎﻥ ﺍﳌﺘﺪﺧﻠﺘﺎﻥ ﳘﺎ:
Ag + / Agﻭ )(s
CH 3COO − / CH 3 CHO
) ( aq
) ( aq
) ( aq
Ag ) × 2
ﺍﺧﺘﺰﺍﻝ e −
( Ag + + aq
) (s −
CH 3CHO + 3 OH → CH 3 COO − + 2 H 2 O + 2e − ) (l
) ( aq
) ( aq
) ( aq
−
2 Ag + CH 3 CHO + 3 OH → 2 Ag + CH 3COO − + 2 H 2 O ) (l
) ( aq
)(s
+
) ( aq
) ( aq
) ( aq
(IIIﺑﻌﺾ ﺗﻄﺒﻴﻘﺎﺕ ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ. ﻳﺘﻢ ﺗﺴﺮﻳﻊ ﺍﻭ ﲣﻔﻴﺾ ﺳﺮﻋﺔ ﲢﻮﻝ ﻛﻴﻤﻴﺎﺋﻲ ﺑﺎﻟﺘﺄﺛﲑ ﻋﻠﻰ ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ. ﻓﺎﻟﺘﺴﺮﻳﻊ ﻗﺪ ﳓﺼﻞ ﻋﻠﻴﻪ ﺇﻣﺎ ﺑﺮﻓﻊ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻮﺳﻂ ﺍﻟﺘﻔﺎﻋﻠﻲ ﺃﻭ ﺑﺎﻟﺰﻳﺎﺩﺓ ﰲ ﺍﻟﺘﺮﻛﻴﺰ ﺍﻟﺒﺪﺋﻲ ﻟﻠﻤﺘﻔﺎﻋﻼﺕ .ﺑﻴﻨﻤﺎ ﺍﻟﺘﺨﻔﻴﺾ ﻣﻦ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺃﻭ ﺍﻟﺘﻮﻗﻴﻒ ﳓﺼﻞ ﻋﻠﻴﻬﻤﺎ ﲞﻔﺾ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﺃﻭ ﺑﺘﺨﻔﻴﻒ ﺍﻟﻮﺳﻂ ﺍﻟﺘﻔﺎﻋﻠﻲ . Abdelkrim SBIRO )(pour toutes observation contactez mon émail [email protected] 6
ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺴﺮﻳﻌﺔ ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺒﻄﻴﺌﺔ Iﺗﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ -ﺍﺧﺘﺰﺍﻝ: (1ﺍﻹﺑﺮﺍﺯ ﺍﻟﺘﺠﺮﻳﱯ ﳌﻔﻬﻮﻡ ﺍﻷﻛﺴﺪﺓ -ﺍﺧﺘﺰﺍﻝ: ﺃ(ﲡﺮﺑﺔ:
ﻧﻐﻤﺮ ﺻﻔﻴﺤﺔ ﻣﻦ ﺍﻟﻨﺤﺎﺱ ﰲ ﻛﺄﺱ ﺑﻪ ﳏﻠﻮﻝ ﻣﺎﺋﻲ ﻟﻨﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ) ( Ag + + NO3 − aq
aq
ﻧﻼﺣﻆ ﺗﻮﺿﻊ ﻃﺒﻘﺔ ﻣﻦ ﺍﻟﻔﻀﺔ Agﻋﻠﻰ ﺍﳉﺰﺀ ﺍﳌﻐﻤﻮﺭ ﻭﺗﻠﻮﻥ ﺍﶈﻠﻮﻝ ﺑﺎﻟﻠﻮﻥ ﺍﻷﺯﻕ ﻧﺘﻴﺠﺔ ﺗﻜﻮﻥ ﺍﻷﻳﻮﻧﺎﺕ . Cu 2+ ) ( aq
) (s
ﺏ(ﺗﻌﻠﻴﻞ: ⊗ ﺧﻼﻝ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺗﺄﻛﺴﺪﺕ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ ﺇﱃ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻨﺤﺎﺱ ﺣﺴﺐ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ : 2e −
+
Cu 2+ ) ( aq
ﺃﻛﺴﺪﺓ
Cu ) (s
ﻭﻫﻜﺬﺍ ﻟﻌﺒﺖ ﺫﺭﺍﺕ ﺍﻟﻨﺤﺎﺱ ﺩﻭﺭ ﺍﳌﺨﺘﺰﻝ)ﻷﺎ ﻓﻘﺪﺕ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ( ﻓﺘﺤﻮﻟﺖ ﺇﱃ ﺍﳌﺆﻛﺴﺪ ﺍﳌﺮﺍﻓﻖ Cu 2+ﻭﻫﺬﺍﻥ ﺍﻟﻨﻮﻋﺎﻥ ) ( aq
ﻳﻨﺘﻤﻴﺎﻥ ﺇﱃ ﺍﳌﺰﺩﻭﺝ . Cu 2+ / Cu )(s
aq
⊗ ﺑﻴﻨﻤﺎ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻔﻀﺔ ﺍﺧﺘﺰﻟﺖ ﺇﱃ Agﺣﺴﺐ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ : ) (s
Ag
ﺍﺧﺘﺰﺍﻝ
e−
) (s
+
+
Ag
) (aq
ﲝﻴﺚ ﻟﻌﺒﺖ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻔﻀﺔ Ag +ﺩﻭﺭ ﺍﳌﺆﻛﺴﺪ)ﻷﺎ ﺍﻛﺘﺴﺒﺖ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ( ﻓﺘﺤﻮﻟﺖ ﺇﱃ ﺍﳌﺨﺘﺰﻝ ﺍﳌﺮﺍﻓﻖ Agﻭﻫﺬﺍﻥ ﺍﻟﻨﻮﻋﺎﻥ ﻳﻨﺘﻤﻴﺎﻥ ) (s
) (aq
ﺇﱃ ﺍﳌﺰﺩﻭﺟﺔ . Ag + / Ag ) (s
) (aq
1
⊗ ﻭﳝﻜﻦ ﺍﳊﺼﻮﻝ ﻋﻠﻰ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﲜﻤﻊ ﻧﺼﻔﻲ ﻣﻌﺎﺩﻟﱵ ﺍﻷﻛﺴﺪﺓ –ﺍﺧﺘﺰﺍﻝ : 2e −
Cu 2+
+
ﺍﻛﺴﺪﺓ
) ( aq
Cu ) (s
ﺍﺧﺘﺰﺍﻝ e −
Ag ) × 2
( Ag + + aq
) (s
+ 2 Ag +
Cu 2+
+ 2 Ag
) ( aq
) (s
) (aq
Cu ) (s
ﺩ( ﺍﺳﺘﻨﺘﺎﺝ: ﺍﳌﺆﻛﺴﺪﻫﻮ ﻛﻞ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻗﺎﺩﺭ ﻋﻠﻰ ﺍﻛﺘﺴﺎﺏ ﺍﻟﻜﺘﺮﻭﻥ ﺃﻭ ﺃﻛﺜﺮ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ،ﻭﻳﺴﻤﻰ ﺍﻟﻨﻮﻉ ﺍﻟﻨﺎﺗﺞ ﳐﺘﺰﻻ ﻣﺮﺍﻓﻘﺎ . ﺍﳌﺨﺘﺰﻝ ﻫﻮ ﻛﻞ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻗﺎﺩﺭ ﻋﻠﻰ ﻣﻨﺢ ﺍﻟﻜﺘﺮﻭﻥ ﺃﻭ ﺃﻛﺜﺮ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ،ﻭﻳﺴﻤﻰ ﺍﻟﻨﻮﻉ ﺍﻟﻨﺎﺗﺞ ﻣﺆﻛﺴﺪﺍ ﻣﺮﺍﻓﻘﺎ . ﺍﻷﻛﺴﺪﺓ ﻫﻲ ﻓﻘﺪﺍﻥ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ﻣﻦ ﻃﺮﻑ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻭﺍﻟﻨﻮﻉ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺬﻱ ﺗﻄﺮﺃ ﻋﻠﻴﻪ ﺍﻷﻛﺴﺪﺓ ﻳﺴﻤﻰ ﳐﺘﺰﻻ. ﻭﺍﻹﺧﺘﺰﺍﻝ ﻫﻮ ﺍﻛﺘﺴﺎﺏ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ﻣﻦ ﻃﺮﻑ ﻧﻮﻉ ﻛﻴﻤﻴﺎﺋﻲ ﻭﺍﻟﻨﻮﻉ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﺍﻟﺬﻱ ﻳﻄﺮﺃ ﻋﻠﻴﻪ ﺍﻹﺧﺘﺰﺍﻝ ﻳﺴﻤﻰ ﻣﺆﻛﺴﺪﺍ. ﺑﺼﻔﺔ ﻋﺎﻣﺔ ﻳﺘﻢ ﺗﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﲔ ﻣﺰﺩﻭﺟﺘﲔ ﳐﺘﺰﻝ/ﻣﺆﻛﺴﺪ ox1 / réd1ﻭ ox 2 / réd 2ﲝﻴﺚ ﳛﺪﺙ ﺍﻧﺘﻘﺎﻝ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ( ox1 + n1e − → réd1 ) × n2
ﻣﻦ ﳐﺘﺰﻝ ﺇﺣﺪﺍ ﳘﺎ ﺇﱃﻣﺆﻛﺴﺪ ﺍﻻﺧﺮﻯ
(réd 2 → ox 2 + n 2 e − ) × n1 ←
n 2 ox1 + n1 réd 2 → n 2 réd 1 + n1ox 2
ل :ﻋﻨﺩﻤﺎ ﻨﻐﻤﺭ ﺼﻔﻴﺤﺔ ﻤﻥ ﺍﻷﻝﻤﻨﻴﻭﻡ ﻓﻲ ﻤﺤﻠﻭل ﻤﺎﺌﻲ ﻝﻨﺘﺭﺍﺕ ﺍﻝﻔﻀﺔ ﻨﺤﺼل ﻋﻠﻰ ﺘﻭﻀﻊ ﺍﻝﻔﻀﺔ ﻋﻠﻰ ﺍﻝﺼﻔﻴﺤﺔ ﻭﻨﺒﺭﺯ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻝﺘﻔﺎﻋل ﺘﻜﻭﻥ ﺃﻴﻭﻨﺎﺕ ﺍﻷﻝﻤﻨﻴﻭﻡ .ﺇﺫﻥ ﺍﻝﺘﻔﺎﻋل ﺍﻝﺫﻱ ﺘﻡ ﺒﻴﻥ ﺍﻝﻤﺯﺩﻭﺠﺘﻴﻥ Al 3+ / Al )(s
) ( aq
و Ag + / Agﻫﻮ ﻛﻤﺎ ﻳﻠﻲ: )(s
) ( aq −
Ag ) × 3
+
( Ag + e
Al 3+ + 3e − 3 Ag + Al 3+
Al Al + 3 Ag +
(2ﺇﺛﺒﺎﺕ ﻣﻌﺎﺩﻟﺔ ﺗﻔﺎﻋﻞ ﺍﻟﻸﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ: ﺃ(
ﲡﺮﺑﺔ:ﻧﺼﺐ ﰲ ﻛﺄﺱ ﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﻛﱪﻳﺘﺎﺕ ﺍﳊﺪﻳﺪ
II
) . ( Fe 2+ + SO4 2−
ﰒ ﻧﻀﻴﻒ ﺇﻟﻴﻪ ﻗﻄﺮﺓ ،ﻗﻄﺮﺓ ،ﳏﻠﻮﻝ ﺑﺮﻣﻐﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ) ( K + + MnO4 −ﺍﶈﻤﺾ ﲝﻤﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ.
ﻧﻼﺣﻆ ﺃﻥ ﳏﻠﻮﻝ ﺑﺮﻣﻐﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ﻳﻔﻘﺪ ﻟﻮﻧﻪ ﺍﻟﺒﻨﻔﺴﺠﻲ ﻧﺘﻴﺠﺔ ﺗﻜﻮﻥ ﺃﻳﻮﻧﺎﺕ ﺍﳌﻨﻐﻨﻴﺰ Mn 2+ﺍﻟﻌﺪﳝﺔ ﺍﻟﻠﻮﻥ ﰲ ﺍﶈﺎﻟﻴﻞ ) ( aq
ﺍﳌﺎﺋﻴﺔ. ﻭ ﻧﱪﺯ ﺑﺎﺳﺘﻌﻤﺎﻝ ﳏﻠﻮﻝ ﺍﻟﺼﻮﺩﺍ ﻭﺟﻮﺩ ﺍﻷﻳﻮﻧﺎﺕ Fe 3+ﰲ ﺍﶈﻠﻮﻝ ﺍﶈﺼﻞ ﻋﻠﻴﻪ)ﺍﻟﱵ ﺗﻌﻄﻲ ﻟﻠﻤﺤﻠﻮﻝ ﻟﻮﻥ ﺍﻟﺼﺪﺃ(. ) ( aq
ﺧﻼﻝ ﻫﺬﻩ ﺍﻟﺘﺠﺮﺑﺔ ﰎ ﺗﻔﺎﻋﻞ ﺑﲔ ﺃﻳﻮﻧﺎﺕ ﺍﻟﱪﻣﻨﻐﻨﺎﺕ MnO4 −ﻭﺃﻳﻮﻧﺎﺕ
ﺍﳊﺪﻳﺪII
) ( aq
ﺍﳊﺪﻳﺪ III
. Fe 3+ ) ( aq
ﺏ( ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ: ﻟﻨﻜﺘﺐ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﻭﺟﺔ
: Fe 3+ / Fe 2+ ) ( aq
) ( aq
2
2+
Feﻹﻋﻄﺎﺀ ﺃﻳﻮﻧﺎﺕ ﺍﳌﻨﻐﻨﻴﺰ
) ( aq
2+
Mnﻭﺃﻳﻮﻧﺎﺕ
) ( aq
Fe 3+ + e −
Fe 2+
) ( aq
−
/ Mn 2+
ﻭﻟﻨﻜﺘﺐ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﻭﺟﺔ
) ( aq
) ( aq
:
MnO4 ) ( aq
ﺑﺼﻔﺔ ﻋﺎﻣﺔ ﳌﻮﺍﺯﻧﺔ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ ﺃﻭ ﺍﻹﺧﺘﺰﺍﻝ ﻧﺘﺘﺒﻊ ﺍﳋﻄﻮﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ : ﻧﻮﺍﺯﻥ ﺍﻟﻌﻨﺼﺮ ﺍﻟﺬﻱ ﺗﻌﺮﺽ ﻟﺘﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ ﺃﻭ ﺍﻹﺧﺘﺰﺍﻝ. ﻧﻮﺍﺯﻥ ﻋﻨﺼﺮ ﺍﻷﻭﻛﺴﺠﲔ ﺑﺈﺿﺎﻓﺔ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ . H 2 O ﻧﻮﺍﺯﻥ ﻋﻨﺼﺮ ﺍﳍﻴﺪﺭﻭﺟﲔ ﺑﺈﺿﺎﻓﺔ ﺍﻟﱪﻭﺗﻮﻧﺎﺕ . H + ﳓﻘﻖ ﻣﻮﺍﺯﻧﺔ ﺍﳌﻌﺎﺩﻟﺔ ﻣﻦ ﺣﻴﺚ ﺍﻟﺸﺤﻨﺎﺕ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ﺑﺈﺿﺎﻓﺔ ﺍﻹﻟﻜﺘﺮﻭﻧﺎﺕ ﺟﻬﺔ ﺍﳌﺆﻛﺴﺪ.ﻓﻨﺤﺼﻞ ﻋﻠﻰ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: + 4 H 2O )(l
−
MnO4 + 8 H + + 5e −
Mn 2+ +
) ( aq
) ( aq
) ( aq
ﻭﳓﺼﻞ ﻋﻠﻰ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﺈﺿﺎﻓﺔ ﻧﺼﻔﻲ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ .
ﺗﻄﺒﻴﻖ:ﺍﻛﺘﺐ ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ -ﺍﺧﺘﺰﺍﻝ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ / Cr 3+
ﺍﳌﺮﺣﻠﺔ ﺍﻷﻭﱃ:
2 Cr 3+
ﺍﳌﺮﺣﻠﺔ ﺍﻟﺜﺎﻧﻴﺔ:
7H 2O
2−
Cr2 O7
2−
Cr2 O7
2 Cr 3+
+
+ 14 H + + 6e −
2−
Cr2 O7
IIﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺴﺮﻳﻌﺔ ﻭﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺒﻄﻴﺌﺔ: (1ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺴﺮﻳﻌﺔ: ﺃ( ﺗﻌﺮﻳﻒ: ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺴﺮﻳﻌﺔ ﻫﻲ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﱵ ﲢﺪﺙ ﰲ ﻣﺪﺓ ﺯﻣﻨﻴﺔ ﻭﺟﻴﺰﺓ ﲝﻴﺚ ﻻ ﳝﻜﻦ ﺗﺘﺒﻊ ﺗﻄﻮﺭﻫﺎ ﺑﺎﻟﻌﲔ ﺍﺮﺩﺓ ﺃﻭ ﺑﺄﺩﻭﺍﺕ ﺍﻟﻘﻴﺎﺱ .
ﺏ( ﺃﻣﺜﻠﺔ: ∗ ﻣﺜﺎﻝ:1ﺗﺮﺳﻴﺐ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﻨﺤﺎﺱ + 2 OH −
Cu (OH ) 2
) ( aq
) ( aq
C u 2+ ) ( aq
∗ ﻣﺜﺎﻝ :2ﺗﺮﺳﻴﺐ ﻛﻠﻮﺭﻭﺭ ﺍﻟﻔﻀﺔ + Cl −
) ( AgCl
) (aq
)(s
Ag + ) (aq
∗ ﻣﺜﺎﻝ:3 ﺍﻟﺘﻔﺎﻋﻞ ﺑﲔ ﺍﳌﺰﺩﻭﺟﺘﲔ / S 2 O3 2− : ) ( aq
2−
S 4 O6
ﻭ
) ( aq
. I2 / I −
) (aq
) ( aq
ﻧﺼﺐ ﰲ ﻛﺄﺱ ﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﺛﻴﻮ ﻛﱪﻳﺘﺎﺕ ﺍﻟﺼﻮﺩﻳﻮﻡ ) (2 Na + + S 2 O3 2−ﻧﻀﻴﻒ ﺇﻟﻴﻪ ﺗﺪﺭﳚﻴﺎ ﳏﻠﻮﻝ ﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ . I 2 ) ( aq
3
) ( aq
) ( aq
ﻳﺪﻝ ﺍﻹﺧﺘﻔﺎﺀ ﺍﻟﻠﺤﻈﻲ ﻟﻠﻮﻥ ﺍﻟﺒﲏ ﻟﺜﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ ﻋﻠﻰ ﺃﻥ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺳﺮﻳﻊ. ﺧﻼﻝ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺗﺄﻛﺴﺪﺕ ﺃﻳﻮﻧﺎﺕ ﺛﻴﻮ ﻛﱪﻳﺘﺎﺕ S 2 O3 2−ﺇﱃ ﺃﻳﻮﻧﺎﺕ ﺛﻴﻮﻧﺎﺕ S 4 O6 2−ﻭﻓﻖ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: ) ( aq 2−
+ 2e −
) ( aq 2−
ﺃﻛﺴﺪﺓ
S 4 O6 ) ( aq
2 S 2 O3 ) ( aq
ﺑﻴﻨﻤﺎ ﺍﺧﺘﺰﻟﺖ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻴﻮﺩ I 2ﺇﱃ I −ﻭﻓﻖ ﻧﺼﻒ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ: ) ( aq ) ( aq
−
−
ﺍﺧﺘﺰﺍﻝ
2 I
) ( aq
+ 2e
I2 ) ( aq
ﻭﳓﺼﻞ ﻋﻠﻰ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﺈﺿﺎﻓﺔ ﻧﺼﻔﻲ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ ﻃﺮﻓﺎ ﺑﻄﺮﻑ. 2 I−
) ( aq
2−
+
2 S 2 O3 2− + I 2
S 4 O6 ) ( aq
) ( aq
) ( aq
(2ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺒﻄﻴﺌﺔ: ﺃ( ﺗﻌﺮﻳﻒ: ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﺒﻄﻴﺌﺔ ﻫﻲ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﱵ ﺗﺴﺘﻐﺮﻕ ﻣﺪﺓ ﺯﻣﻨﻴﺔ ﻧﺴﺘﻄﻴﻊ ﺧﻼﳍﺎ ﺗﺘﺒﻌﻪ ﺑﺎﻟﻌﲔ ﺍﺮﺩﺓ ﺃﻭ ﺑﺄﺩﻭﺍﺕ ﺍﻟﻘﻴﺎﺱ .
ﺏ(ﻣﺜﺎﻝ:1 ﳕﺰﺝ ﰲ ﻛﺄﺱ ﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﻳﻮﺩﻭﺭ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ) ( K + + I −ﻭﺍﳌﺎﺀ ﺍﻷﻛﺴﻴﺠﻴﲏ H 2 O2ﺍﶈﻤﺾ ﺑﻘﻄﺮﺍﺕ ﻣﻦ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ﺍﳌﺮﻛﺰ . ﻣﺒﺎﺷﺮﺓ ﺑﻌﺪ ﻣﺰﺝ ﺍﶈﻠﻮﻟﲔ ﳛﺪﺙ ﺗﻔﺎﻋﻞ ﺑﲔ I −ﻭ H 2 O2ﻓﻴﺘﻜﻮﻥ ﺗﺪﺭﳚﻴﺎ ﺍﳌﺎﺀ ﻭﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ I 2ﺫﻱ ﺍﻟﻠﻮﻥ ﺍﻟﺒﲏ. H 2 O 2 + 2 H + + 2e − → 2 H 2 O
ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ : H 2 O2 / H 2 O ) (l
ﻭﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ
) ( aq
) ( aq
) (l
: I2 / I −
) ( aq
2 I − → I 2 + 2e −
) ( aq ) ( aq
) ( aq
) ( aq
ﻭﳓﺼﻞ ﻋﻠﻰ ﻣﻌﺎﺩﻟﺔ ﺍﻷﻛﺴﺪﺓ-ﺍﺧﺘﺰﺍﻝ ﺑﺈﺿﺎﻓﺔ ﻧﺼﻔﻲ ﺍﳌﻌﺎﺩﻟﺘﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ ﻃﺮﻓﺎ ﺑﻄﺮﻑ. H 2 O2 + 2 I − + 2 H + → I 2 + 2 H 2 O ) (l
) ( aq
) ( aq
ﳝﻜﻦ ﺗﺘﺒﻊ ﺗﻄﻮﺭ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﲟﻌﺎﻳﺮﺓ ﺛﻨﺎﺋﻲ ﺍﻟﻴﻮﺩ I 2ﺍﳌﻜﻮﻥ ﺑﻮﺍﺳﻄﺔ ﺃﻳﻮﻧﺎﺕ ﺛﻴﻮﻛﱪﻳﺘﺎﺕ
) ( aq
2−
) ( aq
. S 2 O3 ) ( aq
) ( aq
ﺝ( ﻣﺜﺎﻝ:2 ﳕﺰﺝ ﰲ ﻛﺄﺱ 10mlﻣﻦ ﳏﻠﻮﻝ ﲪﺾ ﺍﻟﻜﻠﻮﺭﻳﺪﺭﻳﻚ ) ( H + + Cl −ﺍﳌﺮﻛﺰ) (1mol / lﻭﻗﻠﻴﻼ ﻣﻦ ﳏﻠﻮﻝ ﺛﻴﻮ ﻛﱪﻳﺘﺎﺕ ) ( aq
) ( aq
ﺍﻟﺼﻮﺩﻳﻮﻡ ) (2 Na + + S 2 O3 2−ﺫﻱ ﺍﻟﺘﺮﻛﻴﺰ . 10 −1 mol / l ) ( aq
) ( aq
ﻧﻼﺣﻆ ﺃﻥ ﳏﺘﻮﻯ ﺍﻟﻜﺄﺱ ﻳﺄﺧﺬ ﺑﻌﺪ ﳊﻈﺎﺕ ﻟﻮﻧﺎ ﺃﺻﻔﺮﺍﻧﺘﻴﺠﺔ ﺗﻜﻮﻥ ﺍﻟﻜﱪﻳﺖ ﻭﻫﺬﺍ ﺍﻟﺘﺤﻮﻝ ﺑﻄﻲﺀ ﻷﻧﻪ ﻳﺴﺘﻐﺮﻕ ﻣﺪﺓ ﺗﺴﻤﺢ ﺑﺘﺘﺒﻌﻪ ،ﺧﺎﺻﺔ ﻋﻨﺪﻣﺎ ﻧﺴﻠﻂ ﻋﻠﻴﻪ ﺣﺰﻣﺔ ﻣﻦ ﺍﻟﻀﻮﺀ ﺍﻷﺑﻴﺾ ﻭﺩﻗﺎﺋﻖ ﺍﻟﻜﱪﻳﺖ ﺍﻟﻌﺎﻟﻘﺔ ﰲ ﺍﶈﻠﻮﻝ ﺗﺸﺘﺖ ﺍﻟﻀﻮﺀ ﻭﺧﺎﺻﺔ ﺍﳌﻮﺟﺎﺕ ﺍﳌﻮﺍﻓﻘﺔ ﻟﻠﻮﻥ ﺍﻷﺯﺭﻕ ﻓﻴﺄﺧﺬ ﺍﳋﻠﻴﻂ ﰲ ﺍﻟﺒﺪﺍﻳﺔ ﻟﻮﻧﺎ ﺃﺯﺭﻗﺎ ﰒ ﻳﺘﺤﻮﻝ ﺇﱃ ﺍﻷﺻﻔﺮ.
4
ﻳﺘﻤﻴﺰ ﻫﺬﺍ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﺯﺩﻭﺍﺟﻴﺔ ﲢﻮﻝ ﺍﻷﻳﻮﻧﺎﺕ S 2 O3 2−ﰲ ﻭﺳﻂ ﲪﻀﻲ ﻋﻠﻰ ﺇﱃ Sﰒ ﺇﱃ ) . ( SO ) (S
) ( aq
: S 2 O3 2− / S
ﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ + 3 H 2O
S
) (l
) (S
) ( aq
+ 4e −
ﺍﺧﺘﺰﺍﻝ
2
) (S
2−
ﻭﻧﺼﻒ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺰﺩﻭﺟﺔ / S 2 O3 : 4e −
2H + + ) ( aq
) ( aq
) ( aq
SO
ﺃﻛﺴﺪﺓ
2 SO
+ 6H +
2−
S 2 O3
2 ) ( aq
) ( aq
+
2 ) ( aq
+
H 2O
2 ) ( aq
2−
) (l
S 2 O3 ) ( aq
ﻭﺣﺼﻴﻠﺔ ﺍﻟﺘﻔﺎﻋﻞ ﻫﻲ: + SO
+ H 2O
2 ) ( aq
) (l
2H +
S
) ( aq
) (S
2−
+
S 2 O3 ) ( aq
ﺩ( ﺃﻣﺜﻠﺔ ﺍﺧﺮﻯ ﻟﺘﺤﻮﻻﺕ ﺑﻄﻴﺌﺔ : ﺗﻔﺎﻋﻞ ﺍﻷﺳﺘﺮﺓ.-ﺗﻜﻮﻥ ﺍﻟﺼﺪﺃ ﺍﻟﺬﻱ ﻳﻨﺘﺞ ﻋﻨﻪ ﺗﺂﻛﻞ ﺍﳊﺪﻳﺪ.
ﻣﻠﺤﻮﻇﺔ:1
ﺗﻌﺘﱪ ﺑﻌﺾ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺟﺪ ﺑﻄﻴﺌﺔ ﲝﻴﺚ ﺗﻈﻬﺮ ﻭﻛﺄﺎ ﻻﲢﺪﺙ،ﻧﻘﻮﻝ ﺇﺎ ﺣﺮﻛﻴﺎ ﻣﺘﻮﻗﻔﺔ.
ﻣﺜﺎﻝ :ﲢﻠﻞ ﺍﳌﺎﺀ ﺍﻷﻛﺴﻴﺠﻴﲏ
+ O2 )(g
H 2 O2
H 2O ) (l
) ( aq
ﻣﻠﺤﻮﻇﺔ :2ﻹﺑﺮﺍﺯ ﺍﻟﺘﺤﻮﻻﺕ ﺍﻟﺒﻄﻴﺌﺔ ﻧﺴﺘﻌﻤﻞ ﺑﻌﺾ ﺍﻟﺘﻘﻨﻴﺎﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ،ﻣﺜﻞ:ﺍﺳﺘﻌﻤﺎﻝ ﺍﳌﺎﻧﻮﻣﻴﺘﺮ ﺃﻭﻣﻘﻴﺎﺱ ﺍﳌﻮﺍﺻﻠﺔﺃﻭﺍﺳﺘﻌﻤﺎﻝ ﺟﻬﺎﺯ - pHﻣﻴﺘﺮ.
(IIIﺍﻹﺑﺮﺍﺯ ﺍﻟﺘﺠﺮﻳﱯ ﻟﻠﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ: (1ﺗﻌﺮﻳﻒ: ﻧﺴﻤﻲ ﻋﺎﻣﻼ ﺣﺮﻛﻴﺎ ﻛﻞ ﻣﻘﺪﺍﺭ ﻣﺆﺛﺮ ﻋﻠﻰﺳﺮﻋﺔ ﲢﻮﻝ ﻛﻴﻤﻴﺎﺋﻲ .
(2ﺃﻣﺜﻠﺔ ﻟﺒﻌﺾ ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ :
ﺃ(ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ : ﻧﻨﺠﺰ ﺍﻟﺘﺠﺮﺑﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻹﺑﺮﺍﺯ ﺗﺄﺛﲑ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﻋﻠﻰ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ . ﻧﺼﺐ ﰲ ﺃﻧﺒﻮﰊ ﺇﺧﺘﺒﺎﺭ )(1ﻭ) 10ml (2ﻣﻦ ﳏﻠﻮﻝ ﲪﺾ ﺍﻷﻭﻛﺴﺎﻟﻴﻚ H 2 C 2 O4ﺫﻱ ﺗﺮﻛﻴﺰ . 0,5mol / l ) ( aq
ﻧﺘﺮﻙ ﺍﻟﻜﺄﺱ ) (1ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻋﺎﺩﻳﺔ ﻭﻧﺮﻓﻊ ﺩﺭﺟﺔ ﺍﻟﺜﺎﱐ ﺇﱃ . 40°c ﻧﻀﻴﻒ ﰲ ﻧﻔﺲ ﺍﻟﻠﺤﻈﺔ ﺇﱃ ﻛﻞ ﻣﻨﻬﻤﺎ 5mlﺑﺮﻣﻐﻨﺎﺕ ﺍﻟﺒﻮﺗﺎﺳﻴﻮﻡ ) . ( K + + MnO4 − ﻧﻼﺣﻆ ﺍﺧﺘﻔﺎﺀ ﺍﻟﻠﻮﻥ ﺍﻟﺒﻨﻔﺴﺠﻲ ﺍﳌﻤﻴﺰ ﻟﻸﻳﻮﻧﺎﺕ MnO4 −ﺃﺳﺮﻉ ﻗﻲ ﺍﻟﻜﺄﺱ 2 ﺍﻟﺬﻱ ﰎ ﺗﺴﺨﻴﻨﻪ )ﺃﻱ ﺭﻓﻊ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺗﻪ(. ﻧﺴﺘﻨﺘﺞ ﺃﻧﻪ ﻛﻠﻤﺎ ﻛﺎﻧﺖ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﻣﺮﺗﻔﻌﺔ ﻛﻠﻤﺎ ﺍﺯﺩﺍﺩﺕ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ.
ﺏ( ﺗﺎﺛﲑ ﺗﺮﺍﻛﻴﺰ ﺍﳌﺘﻔﺎﻋﻼﺕ : 5
1
2
ﺗﺘﻔﺎﻋﻞ ﺍﻳﻮﻧﺎﺕ ﺍﻟﻴﻮﺩﻭﺭ I −ﻣﻊ ﺟﺰﻳﺌﺎﺕ ﺍﳌﺎﺀ ﺍﻷﻛﺴﻴﺠﻴﲏ H 2 O2ﰲ ﻭﺳﻂ ﲪﻀﻲ ﺣﺴﺐ ﺍﳌﻌﺎﺩﻟﺔ : H 2 O2 + 2 I − + 2 H + → I 2 + 2 H 2 O ) (l
) ( aq
) ( aq
) ( aq
) ( aq
ﻧﺼﺐ ﰲ ﻛﺄﺳﲔ ﻧﻔﺲ ﺍﳊﺠﻢ H 2 O2ﺍﶈﻤﺾ ﺑﻘﻄﺮﺍﺕ ﻣﻦ ﲪﺾ ﺍﻟﻜﱪﻳﺘﻴﻚ ﻭﻧﻀﻴﻒ ﻟﻜﻞ ﻣﻨﻬﻤﺎ ﻧﻔﺲ ﺍﳊﺠﻢ ﻣﻦ ﳏﻠﻮﻝ ) ( K + + I −ﻟﻜﻦ ﺗﺮﻛﻴﺰ ﻫﺬﺍ ﺍﻻﺧﲑ ﰲ ﺍﻟﻜﺎﺱ ﺍﻟﺜﺎﱐ ﺃﻛﱪ ﻣﻨﻪ ﰲ ﺍﻟﻜﺄﺱ ﺍﻷﻭﻝ . ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻳﻜﻮﻥ ﺃﺳﺮﻉ ﻛﻠﻤﺎ ﻛﺎﻥ ﺗﺮﻛﻴﺰﺍﺣﺪ ﺍﳌﺘﻔﺎﻋﻼﺕ ﺍﻛﱪ. ﻛﻤﺎ ﻧﺸﲑ ﺍﱃ ﺍﻥ ﻫﻨﺎﻙ ﻋﻮﺍﻣﻞ ﺍﺧﺮﻯ ﻣﺜﻞ ﺍﳊﻔﺎﺯ ﻭﺍﳌﺬﻳﺐ .
(3ﺗﻔﺎﻋﻼﻥ ﺧﺎﺻﺎﻥ: ﺃ( ﻣﺘﻔﺎﻋﻞ ﻓﻬﻠﲔ :ﺗﻔﺎﻋﻞ ﺧﺎﺹ ﺑﻄﻲﺀ ﻳﺴﺘﻌﻤﻞ ﻛﺮﺍﺋﺰ ﻟﺘﻤﻴﻴﺰ ﺍﻷﻟﺪﻫﻴﺪﺍﺕ . ﳛﺘﻮﻱ ﳏﻠﻮﻝ ﺍﻟﻔﻬﻠﲔ ﻋﻠﻰ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻨﺤﺎﺱ IIﻣﻌﻘﺪ ﰲ ﻭﺳﻂ ﻗﺎﻋﺪﻱ ﺗﻌﻄﻲ ﻟﻠﻤﺤﻠﻮﻝ ﻟﻮﻧﺎ ﺃﺯﺭﻗﺎ ﺩﺍﻛﻨﺎ،ﻭﻟﻠﺘﺒﺴﻴﻂ ﻧﺮﻣﺰ ﳍﺬﻩ ﺍﻷﻳﻮﻧﺎﺕ ﺏ Cu 2+ ( aq ) :ﻭﻫﻲ ﺍﻟﱵ ﺗﻜﻮﻥ ﻣﺆﻛﺴﺪ ﺍﳌﺰﺩﻭﺟﺔ. Cu 2+ / Cu 2 O : ) ( aq
) ( aq
ﻧﺼﺐ ﰲ ﺃﻧﺒﻮﺏ ﺍﺧﺘﺒﺎﺭ ﻗﻠﻴﻼ ﻣﻦ ﺍﻟﻔﻬﻠﲔ ﻭﻧﻀﻴﻒ ﺇﻟﻴﻬﺎ ﻗﻠﻴﻼ ﻣﻦ ﺍﻹﻳﺜﺎﻧﺎﻝ CH 3CHOﻭﻫﻮ ﺃﻟﺪﻫﻴﺪ. ﺍﻟﺘﻔﺎﻋﻞ ﺍﳊﺎﺻﻞ ﺟﺪ ﺑﻄﻲﺀ ﻟﻜﻦ ﻋﻨﺪ ﺗﺴﺨﻴﻨﻪ ﺑﺒﻂﺀ ﻳﺘﻜﻮﻥ ﺍﻭﻛﺴﻴﺪ ﺍﻟﻨﺤﺎﺱ Cu 2 Oﻋﻠﻰ ﺷﻜﻞ ﺭﺍﺳﺐ ﺫﻱ ﻟﻮﻥ ﺃﲪﺮ ﺁﺟﻮﰊ . ﺇﺫﻥ ﰎ ﺗﻔﺎﻋﻞ ﺑﲔ ﺍﳌﺰﺩﻭﺟﺘﲔ :
CH3COO− / CH3CHO
ﻭ. Cu 2+ / Cu 2 O : ) ( aq
) ( aq
CH 3CHO + 3 OH − → CH 3 COO − + 2 H 2 O + 2e − ) (l
) ( aq
) ( aq −
) ( aq 2+
−
2 Cu + 2 OH + 2e → Cu 2 O + H 2 O ) (l
) ( aq
)(s −
→ CH 3COO + 3 H 2 O + Cu 2 O ) ( aq
) (l
2+
) ( aq
) ( aq
−
CH 3CHO + 5 OH + 2 Cu
) ( aq
ﺏ( ﻣﺘﻔﺎﻋﻞ ﺗﻮﻟﻴﻨﺲ
) ( aq
)ﺍﳌﺴﻤﻰ ﻛﺬﻟﻚ:ﻧﺘﺮﺍﺕ ﺍﻟﻔﻀﺔ ﺍﻷﻣﻮﻧﻴﺎﻛﻲ ( Ag (NH 3 ) 2 +ﺗﻔﺎﻋﻞ ﺁﺧﺮ
ﺧﺎﺹ ﺑﻄﻲﺀ ﻳﺴﺘﻌﻤﻞ ﻛﺮﺍﺋﺰ ﻟﺘﻤﻴﻴﺰ ﺍﻷﻟﺪﻫﻴﺪﺍﺕ . ﻳﺘﻔﺎﻋﻞ ﻫﺬﺍ ﺍﻟﺮﺍﺋﺰ ﻣﻊ ﺍﻹﺛﺎﻧﺎﻝ ﰲ ﻭﺳﻂ ﻗﺎﻋﺪﻱ ﻓﻴﺘﻜﻮﻥ ﺭﺍﺳﺐ ﻣﻦ ﺍﻟﻔﻀﺔ ﺍﻟﺬﻱ ﻳﺴﺘﻌﻤﻞ ﻟﺘﻔﻀﻴﺾ ﺍﳌﺮﺍﻳﺎ . ﺍﳌﺰﺩﻭﺟﺘﺎﻥ ﺍﳌﺘﺪﺧﻠﺘﺎﻥ ﳘﺎ:
Ag + / Agﻭ )(s
CH 3COO − / CH 3 CHO
) ( aq
) ( aq
) ( aq
Ag ) × 2
ﺍﺧﺘﺰﺍﻝ e −
( Ag + + aq
) (s −
CH 3CHO + 3 OH → CH 3 COO − + 2 H 2 O + 2e − ) (l
) ( aq
) ( aq
) ( aq
−
2 Ag + CH 3 CHO + 3 OH → 2 Ag + CH 3COO − + 2 H 2 O ) (l
) ( aq
)(s
+
) ( aq
) ( aq
) ( aq
(IIIﺑﻌﺾ ﺗﻄﺒﻴﻘﺎﺕ ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ. ﻳﺘﻢ ﺗﺴﺮﻳﻊ ﺍﻭ ﲣﻔﻴﺾ ﺳﺮﻋﺔ ﲢﻮﻝ ﻛﻴﻤﻴﺎﺋﻲ ﺑﺎﻟﺘﺄﺛﲑ ﻋﻠﻰ ﺍﻟﻌﻮﺍﻣﻞ ﺍﳊﺮﻛﻴﺔ. ﻓﺎﻟﺘﺴﺮﻳﻊ ﻗﺪ ﳓﺼﻞ ﻋﻠﻴﻪ ﺇﻣﺎ ﺑﺮﻓﻊ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻮﺳﻂ ﺍﻟﺘﻔﺎﻋﻠﻲ ﺃﻭ ﺑﺎﻟﺰﻳﺎﺩﺓ ﰲ ﺍﻟﺘﺮﻛﻴﺰ ﺍﻟﺒﺪﺋﻲ ﻟﻠﻤﺘﻔﺎﻋﻼﺕ .ﺑﻴﻨﻤﺎ ﺍﻟﺘﺨﻔﻴﺾ ﻣﻦ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺃﻭ ﺍﻟﺘﻮﻗﻴﻒ ﳓﺼﻞ ﻋﻠﻴﻬﻤﺎ ﲞﻔﺾ ﺩﺭﺟﺔ ﺍﳊﺮﺍﺭﺓ ﺃﻭ ﺑﺘﺨﻔﻴﻒ ﺍﻟﻮﺳﻂ ﺍﻟﺘﻔﺎﻋﻠﻲ . Abdelkrim SBIRO )(pour toutes observation contactez mon émail [email protected] 6
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