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Redox Reactions Atom 1 Atom 2 Gives electrons This atom Oxidizes itself (loses electrons) It’s the reducing agent This atom Reduces itself (gains electrons)

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Presentation on theme: "Redox Reactions Atom 1 Atom 2 Gives electrons This atom Oxidizes itself (loses electrons) It’s the reducing agent This atom Reduces itself (gains electrons)"— Presentation transcript:

1 Redox Reactions Atom 1 Atom 2 Gives electrons This atom Oxidizes itself (loses electrons) It’s the reducing agent This atom Reduces itself (gains electrons) It’s the oxidizing agent Example: Zn Zn 2+ + is oxidized Cu 2+ + Cu is reduced 2e- Zn + Cu 2+ Cu + Zn 2+ Ion Electron reaction equation balancing (half-cells)

2 2) Separate the equation into two half reactions (oxidation and reduction) 3) Balance the atoms Other than H and O Acidic medium: 1)1) Algorithm to balance Redox reactions equations ion-electron Method (acidic medium) 5) Balance charges adding e- Then multiply reactions by factors to make the e- equal in both half reactions 5) Balance charges adding e- Then multiply reactions by factors to make the e- equal in both half reactions 6) Sum both half reaction and check they are balanced (in elements and charges) 1) Write the unbalanced equation of the reaction in the ionic form 4) Balance O, adding H 2 O and H, adding H +

3 2) Separate the equation into two half reactions (oxidation and reduction) 2) Separate the equation into two half reactions (oxidation and reduction) 3) Balance the atoms Other than H and O 1)1) 1) Write the unbalanced equation of the reaction in the ionic form Algorithm to balance Redox reactions equations ion-electron Method (basic medium) 5) Balance charges adding e -, then multiply reactions by factors to make the e - equal in both half reactions 7) Unite the common terms on one side of the half reactions, and sum them. Check they still are balanced (in elements and charges) The procedure is the same as for Acidic medium until Step (5). Then the method changes for step 6 and 7 of the basic medium 4) Balance O, adding H 2 O and H, adding H + 6)Add an OH - for each H + on any side of the equations, and group OH - and H + into one H 2 O on both sides of each half reaction

4 Balance The following equation, in acid solution: Step # 1 Ionic equation: Cr 2 O I - Cr 3+ + I 2 (acidic medium) Step #2: Separate the equation into two half reactions (oxidation and reduction) Cr 2 O 7 2- Cr 3+ [reduction] I - I 2 [oxidation] Step #3: Balance atoms For the Cr 2 O 7 2- /Cr 3+ half reaction: a) Balance atoms other than O e H. Balance the 2 Cr’s on the left side putting a coefficient of 2 on the right. Cr 2 O Cr 3+

5 Step 4. Balance the O atoms, adding H 2 O molecules on the other side, you must add 7 of them on thr right hand side to balance 7 0’s n the left: Cr 2 O Cr H 2 O Balance the H adding H + on the other side. Two H atoms per water molecule, i.e. 14 H + are added on the left hand side: 14 H + + Cr 2 O Cr H 2 O Step 5. Balance charges adding electrons. On the left side, the charge is +12 and on the right it is +6. We must add 6 electrons to the left hand side of the equation 6 e H + + Cr 2 O Cr H 2 O For the I - /I 2 half reaction: No need to balance non O nor non H. Iodine atoms are balanced by placing a 2 coefficient on the left 2 I - I 2

6 Balance charges adding electrons. To balance the two charges on the left, we need to add two electrons on the right. 2 I - I e - 3(2I - I e - ) 6 I - 3 I e - Step # 6 Sum both half reactions, setting common terms on one side of the equation: 6 e H + + Cr 2 O Cr H 2 O 6 I - 3 I e - 6 I H + + Cr 2 O I H 2 O + 2 Cr 3+ Check: reactives(6I, 14H, 2Cr, 7O, charge 6+) = for products! Set the number of electrons equal by multiplying the half reactions by coefficients

7 Redox Reactions Balancing, Basic Solution Balance the following reaction equation in basic solution. 1) MnO C 2 O 4 2- MnO 2 + CO 3 2- Follow same steps as acidic, till step 5 2) Separate into half reactions: MnO 4 - MnO 2 C 2 O 4 2- CO ) Balancing the reduction half reaction: a) non O nor H atoms b) O atoms by water c) H Atoms by H + d) Charges by e - No need MnO 4 - MnO H 2 O 4 H + + MnO 4 - MnO H 2 O 3 e H + + MnO 4 - MnO H 2 O

8 Balancing the oxidation half reaction C 2 O 4 2- CO 3 2- Balancing: a) non O nor H atoms b) O atoms by water c) H Atoms by H+ d) Charges by e- C 2 O CO H 2 O + C 2 O CO H 2 O + C 2 O CO H + 2 H 2 O + C 2 O CO H e - Multiply each half reaction by a factor such as to equal the electron number in both half reaction: oxidation = 2e -, reduction = 3e -, Therefore (oxid x 3 and red x 2 = 6e - ) oxid = 6 H 2 O + 3 C 2 O CO H e - red = 6 e H MnO MnO H 2 O

9 6) Medium change: Add OH- on both sides of equation for each H+ present 12 OH H 2 O + 3 C 2 O CO H OH e - 6 e - + 8OH H MnO MnO H 2 O + 8OH - 4 OH C 2 O MnO 4 - 6CO MnO H 2 O Where we have H + and OH -, group them as H 2 O : Check: reagents (2Mn, 24O, 6C, 4H, charges: 12-) = products 12 OH H 2 O + 3 C 2 O CO H 2 O + 6 e- 6e- + 8H 2 O + 2 MnO MnO H 2 O + 8OH- Sum and group common terms:


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