OxidationReduction Dr Ron Rusay Balancing OxidationReduction Reactions Copyright
Oxidation-Reduction Dr. Ron Rusay Balancing Oxidation-Reduction Reactions © Copyright 2002 -2010 R. J. Rusay
Oxidation-Reduction � Oxidation is the loss of electrons. � Reduction is the gain of electrons. � The reactions occur together. One does not occur without the other. � The terms are used relative to the change in the oxidation state or oxidation number of the reactant(s). © Copyright 2002 -2010 R. J. Rusay
Oxidation Reduction Reactions
QUESTION
QUESTION What is the oxidation number of chromium in ammonium dichromate? A) +3 B) +4 C) +5 D) +6
Zinc
QUESTION Select all redox reactions by looking for a change in oxidation number as reactants are converted to products. I) Ca + 2 H 2 O → Ca(OH)2 + H 2 II) Ca. O + H 2 O → Ca(OH)2 III) Ca(OH)2 + H 3 PO 4 → Ca 3(PO 4)2 + H 2 O IV) Cl 2 + 2 KBr → Br 2 + 2 KCl A) I and II B) II and III C) I and IV D) III and IV
QUESTION
Number of electrons gained must equal the number of electrons lost. - 2 e+2 e- Use oxidation numbers to determine what is oxidized and what is reduced. 0 +2 e- Cu 2+ 0 Refer to Balancing H 2 Oxidation-Reduction Reactions (g) Cu (s) - 2 e- 2 H+
QUESTION
Balancing Redox Equations in acidic solutions 1) Determine the oxidation numbers of atoms in both reactants and products. 2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”. 3) Balance the “redox” atoms and charges (electron gain and loss must equal!). 4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s). © Copyright 2002 -2010 R. J. Rusay
Balancing Redox Equations Fe+2(aq)+ Cr 2 O 72 -(aq) +H+(aq)----> Fe 3+(aq) + Cr 3+(aq) + H 2 O(l) ? Fe 2+(aq)+ Cr 2 O 72 -(aq) +H+(aq)----> Fe 3+(aq) + Cr 3+(aq) + H 2 O(l) x = ? Cr ; 2 x+7(-2) = -2; x = +6 © Copyright 2002 -2010 R. J. Rusay
Balancing Redox Equations Fe 2+(aq)-e - ---> Fe 3+(aq) Cr 2 O 72 -(aq) + 6 e - --> 2 Cr 3+(aq) Cr = (6+) 6 (Fe 2+(aq) -e - ---> Fe 3+(aq)) 6 Fe 2+(aq) ---> 6 Fe 3+(aq) + 6 e Cr 2 O 72 -(aq) + 6 e - --> 2 Cr 3+(aq) © Copyright 2002 -2010 R. J. Rusay
Balancing Redox Equations 6 Fe 2+(aq) ---> 6 Fe 3+(aq) + 6 e Cr 2 O 72 -(aq) + 6 e - --> 2 Cr 3+(aq) 6 Fe 2+(aq)+ Cr 2 O 72 -(aq) + ? 2 nd H+(aq) ----> 6 Fe 3+(aq) + 2 Cr 3+(aq)+ ? 1 st Oxygen H 2 O(l) Oxygen =7 © Copyright 2002 -2010 R. J. Rusay 2 nd (Hydrogen) = 14
Balancing Redox Equations Completely Balanced Equation: 6 Fe 2+(aq)+ Cr 2 O 72 -(aq) + 14 H+(aq) ----> 6 Fe 3+(aq) + 2 Cr 3+(aq)+ 7 H 2 O(l) © Copyright 2002 -2010 R. J. Rusay
QUESTION Dichromate ion in acidic medium converts ethanol, C 2 H 5 OH, to CO 2 according to the unbalanced equation: Cr 2 O 72−(aq) + C 2 H 5 OH(aq) → Cr 3+(aq) + CO 2(g) + H 2 O(l) The coefficient for H+ in the balanced equation using smallest integer coefficients is: A) 8 B) 10 C) 13 D) 16
Balancing Redox Equations in basic solutions 1) Determine oxidation numbers of atoms in Reactants and Products 2) Identify and select out those which change oxidation number into separate “half reactions” 3) Balance redox atoms and charges (electron gain and loss must equal!) 4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water © Copyright 2002 -2010 R. J. Rusay
Balancing Redox Equations in basic solutions Mn. O 2 (aq)+ Cl. O 31 -(aq) + OH 1 -aq) -> Mn. O 41 - (aq)+ Cl 1 -(aq) + H 2 O(l) Mn 4+ (Mn. O 2) ---> Mn 7+ (Mn. O 4 ) 1 Cl+5 (Cl. O 3 ) 1 -+ 6 e- ---> Cl 1 - © Copyright 2002 -2010 R. J. Rusay
Balancing Redox Equations in basic solutions Electronically Balanced Equation: 2 Mn. O 2 (aq)+ Cl. O 31 -(aq) + 6 e - ----> 2 Mn. O 4 1 - + Cl 1 - + 6 e- © Copyright 2002 -2010 R. J. Rusay
Balancing Redox Equations in basic solutions Completely Balanced Equation: 2 Mn. O 2 (aq)+ Cl. O 31 -(aq) + 2 OH 1 - (aq)----> 2 Mn. O 4 (aq)1 - + Cl 1 - (aq)+ 1 H 2 O (l) 9 O in product © Copyright 2002 -2010 R. J. Rusay
QUESTION Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how Mn. O 4– could be used to analyze samples for oxalate. Mn. O 4– + C 2 O 42– Mn. O 2 + CO 32– (basic solution) When properly balanced, how many OH– are present? • • 1 2 3 4
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