Reazioni redox in forma molecolare difficili - 7

TA KIMIKA - esercizi risolti
	Esercizi
	Reazioni redox in forma molecolare difficili
	Teoria - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10
	Soluzione con metodo differente: semireazione \| ione-elettrone

Bilanciare la seguente reazione redox: K₄[Fe(CN)₆] + KMnO₄ + H₂SO₄ => KHSO₄ + MnSO₄ + HNO₃ + CO₂ + Fe₂(SO₄)₃ + H₂O.
Fonte: http://www.webalice.it/fiorema/Notizie/5A_Liceo_S_T_/REAZIONI_REDOX_IMPEGNATIVE.pdf

Punto 1 - Determinare il numero di ossidazione degli atomi presenti
+1 +2 +2-3 +1+7 -2 +1 +6-2 +1+1+6-2 +2 +6-2 +1+5-2 +4-2 +3 +6 -2 +1 -2 K₄[Fe(CN)₆] + KMnO₄ + H₂SO₄ => KHSO₄ + MnSO₄ + HNO₃ + CO₂ + Fe₂(SO₄)₃ + H₂O

Punto 2 - Collegare gli atomi con variazione di n.o. con delle frecce
+1 +2 +2-3 +1+7 -2 +1 +6-2 +1+1+6-2 +2 +6-2 +1+5-2 +4-2 +3 +6 -2 +1 -2 K₄[Fe(CN)₆] + KMnO₄ + H₂SO₄ => KHSO₄ + MnSO₄ + HNO₃ + CO₂ + Fe₂(SO₄)₃ + H₂O \| \| \| \|________________________________________________\| \| \| \| \| \| \| \| \| \| \| \| \|________________________________________________________________________________________\| \| \| \| \|_________________________________________________________________________________________________________\| \| \|____________________________________________________________________________________________________________________________\|

Punto 2 - Collegare gli atomi con variazione di n.o. con delle frecce

    +1    +2  +2-3                 +1+7 -2       +1 +6-2         +1+1+6-2             +2 +6-2          +1+5-2          +4-2          +3    +6 -2          +1 -2  
  K₄[Fe(CN)₆] +    KMnO₄ + H₂SO₄ => KHSO₄ +    MnSO₄ +   HNO₃ +   CO₂ +  Fe₂(SO₄)₃ + H₂O
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Le semireazioni di ossidazione sono tre e tutte legate ad atomi facenti parte del composto K₄[Fe(CN)₆]. Il bilanciamento del numero di atomi di Fe comporta un coefficiente 2 davanti al composto K₄[Fe(CN)₆] e da ciò seguono i bilanciamenti degli atomi di C ed N.

Punto 3 - Bilanciare gli atomi collegati dalle frecce
+1 +2 +2-3 +1+7 -2 +1 +6-2 +1+1+6-2 +2 +6-2 +1+5-2 +4-2 +3 +6 -2 +1 -2 2K₄[Fe(CN)₆] + KMnO₄ + H₂SO₄ => KHSO₄ + MnSO₄ + 12HNO₃ + 12CO₂ + Fe₂(SO₄)₃ + H₂O \| \| \| \|________________________________________________\| \| \| \| \| \| \| \| \| \| \| \| \|________________________________________________________________________________________\| \| \| \| \|_________________________________________________________________________________________________________\| \| \|____________________________________________________________________________________________________________________________\|

Punto 3 - Bilanciare gli atomi collegati dalle frecce

    +1    +2  +2-3                 +1+7 -2       +1 +6-2         +1+1+6-2             +2 +6-2          +1+5-2          +4-2          +3    +6 -2          +1 -2  
 2K₄[Fe(CN)₆] +    KMnO₄ + H₂SO₄ => KHSO₄ +    MnSO₄ + 12HNO₃ + 12CO₂ +  Fe₂(SO₄)₃ + H₂O
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Punto 4 - Esprimere con e^- sulle frecce le 2 variazioni complessive di n.o.
+1 +2 +2-3 +1+7 -2 +1 +6-2 +1+1+6-2 +2 +6-2 +1+5-2 +4-2 +3 +6 -2 +1 -2 2K₄[Fe(CN)₆] + KMnO₄ + H₂SO₄ => KHSO₄ + MnSO₄ + 12HNO₃ + 12CO₂ + Fe₂(SO₄)₃ + H₂O \| \| \| \|__________5e____________________________________\| \| \| \| \| \| \| \| \| \| \| \| \|______________________________96e_______________________________________________________\| \| \| \| \|________________________________24e______________________________________________________________________\| tot=122e \| \|______________________________________2e____________________________________________________________________________________\|

Punto 4 - Esprimere con e^- sulle frecce le 2 variazioni complessive di n.o.

    +1    +2  +2-3                 +1+7 -2       +1 +6-2         +1+1+6-2             +2 +6-2          +1+5-2          +4-2          +3    +6 -2          +1 -2  
 2K₄[Fe(CN)₆] +    KMnO₄ + H₂SO₄ => KHSO₄ +    MnSO₄ + 12HNO₃ + 12CO₂ +  Fe₂(SO₄)₃ + H₂O
          |    | |                    |__________5e____________________________________|                  |              |             |
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          |    | |______________________________96e_______________________________________________________|              |             |
          |    |________________________________24e______________________________________________________________________| tot=122e    |
          |______________________________________2e____________________________________________________________________________________|

Anche in questo caso gli elettroni messi in gioco possono essere computati nel loro complesso.

Punto 5 - Bilanciare gli elettroni ed eseguire i prodotti
+1 +2 +2-3 +1+7 -2 +1 +6-2 +1+1+6-2 +2 +6-2 +1+5-2 +4-2 +3 +6 -2 +1 -2 2K₄[Fe(CN)₆] + KMnO₄ + H₂SO₄ => KHSO₄ + MnSO₄ + 12HNO₃ + 12CO₂ + Fe₂(SO₄)₃ + H₂O \| \| \| \|__________5ex122=610e___________________________\| \| \| \| \| \| \| \| \| \| \| \| \|______________________________96ex5=480e________________________________________________\| \| \| \| \|________________________________24ex5=120e_______________________________________________________________\| tot=610e \| \|______________________________________2ex5=10e______________________________________________________________________________\|
+1 +2 +2-3 +1+7 -2 +1 +6-2 +1+1+6-2 +2 +6-2 +1+5-2 +4-2 +3 +6 -2 +1 -2 10K₄[Fe(CN)₆] + 122KMnO₄ + H₂SO₄ => KHSO₄ + 122MnSO₄ + 60HNO₃ + 60CO₂ + 5Fe₂(SO₄)₃ + H₂O \| \| \| \|__________5ex122=610e___________________________\| \| \| \| \| \| \| \| \| \| \| \| \|______________________________96ex5=480e________________________________________________\| \| \| \| \|________________________________24ex5=120e_______________________________________________________________\| tot=610e \| \|______________________________________2ex5=10e______________________________________________________________________________\|

Punto 5 - Bilanciare gli elettroni ed eseguire i prodotti

    +1    +2  +2-3                 +1+7 -2       +1 +6-2         +1+1+6-2             +2 +6-2          +1+5-2          +4-2          +3    +6 -2          +1 -2  
 2K₄[Fe(CN)₆] +    KMnO₄ + H₂SO₄ => KHSO₄ +    MnSO₄ + 12HNO₃ + 12CO₂ +  Fe₂(SO₄)₃ + H₂O
          |    | |                    |__________5ex122=610e___________________________|                  |              |             |
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          |    | |______________________________96ex5=480e________________________________________________|              |             |
          |    |________________________________24ex5=120e_______________________________________________________________| tot=610e    |
          |______________________________________2ex5=10e______________________________________________________________________________|

    +1    +2  +2-3                 +1+7 -2       +1 +6-2         +1+1+6-2             +2 +6-2          +1+5-2          +4-2          +3    +6 -2          +1 -2  
10K₄[Fe(CN)₆] + 122KMnO₄ + H₂SO₄ => KHSO₄ + 122MnSO₄ + 60HNO₃ + 60CO₂ + 5Fe₂(SO₄)₃ + H₂O
          |    | |                    |__________5ex122=610e___________________________|                  |              |             |
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          |    | |______________________________96ex5=480e________________________________________________|              |             |
          |    |________________________________24ex5=120e_______________________________________________________________| tot=610e    |
          |______________________________________2ex5=10e______________________________________________________________________________|

Punto 6 - Bilanciare la massa
10K₄[Fe(CN)₆] + 122KMnO₄ + H₂SO₄ => 162KHSO₄ + 122MnSO₄ + 60HNO₃ + 60CO₂ + 5Fe₂(SO₄)₃ + H₂O Bilanciamento K
10K₄[Fe(CN)₆] + 122KMnO₄ + 299H₂SO₄ => 162KHSO₄ + 122MnSO₄ + 60HNO₃ + 60CO₂ + 5Fe₂(SO₄)₃ + H₂O Bilanciamento SO₄^2-
10K₄[Fe(CN)₆] + 122KMnO₄ + 299H₂SO₄ => 162KHSO₄ + 122MnSO₄ + 60HNO₃ + 60CO₂ + 5Fe₂(SO₄)₃ + 188H₂O Bilanciamento H e O
10K₄[Fe(CN)₆] + 122KMnO₄ + 299H₂SO₄ => 162KHSO₄ + 122MnSO₄ + 60HNO₃ + 60CO₂ + 5Fe₂(SO₄)₃ + 188H₂O