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Redox Reactions

Redox Reactions. Redox is short for “ oxidation-reduction .” Plating and combustion are redox reactions. Consider the following redox reaction: Fe (s) + Ni(NO 3 ) 2 (aq)  Fe(NO 3 ) 2 (aq) + Ni (s). O xidation I s L oss of electrons R eduction I s G ain of electrons.

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Redox Reactions

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  1. Redox Reactions • Redox is short for “oxidation-reduction.” • Plating and combustion are redox reactions. • Consider the following redox reaction: • Fe(s) + Ni(NO3)2 (aq) Fe(NO3)2(aq)+ Ni (s) Oxidation Is Loss of electrons Reduction Is Gain of electrons • Fe goes from an element to an ion: Fe  Fe2+ • Fe got oxidized • Ni goes from an ion to an element: Ni2+  Ni • Ni2+ got reduced

  2. Oxidation Numbers • We determine whether a chemical species has been oxidized or reduced by looking at the change in its oxidation number. • Fe(s) + Ni(NO3)2 (aq) Fe(NO3)2(aq)+ Ni (s) • ox # 0 +2 +2 0 • The oxidation number of an element is 0. • The oxidation number of a monatomic ion is its charge. • The ox # for Fe went up. It was oxidized. • The ox # for Ni2+ went down. It was reduced (reduced = down). Oxidation Is Loss of electrons Reduction Is Gain of electrons

  3. Oxidation Numbers You are responsible for knowing the rules for assigning oxidation numbers to elements in ions and compounds.

  4. Oxidation Numbers • The oxidation number of an element is 0. • The oxidation number of a monatomic ion is its charge. • The oxidation number of oxygen in a compound is usually -2 (exception is peroxide, O22-, where the oxidation number is -1). • The oxidation number of hydrogen in a compound is 1 when bonded to nonmetals and -1 when bonded to metals. • The oxidation number of fluoride is always -1. • The sum of the oxidation numbers in a neutral compound is zero. • The sum of the oxidation numbers in a polyatomic ion is the charge of the ion.

  5. Oxidation Numbers - Examples Species oxidation Species oxidation number number Fe Na+ H2 H+ O2 O2- Al3+ Cl- Fe2+ N3- S in SO42- : Cr in Cr2O72- N in NO3- S in SO32- N in NO2- P in PO43- C in CO32- 0 0 0 +3 +2 +1 +1 -2 -1 -3 overall charge of ion is -2 = ox # of S + 4(ox # of O) -2 = ox # of S + 4(-2) -2 = ox # of S – 8 -2+8 = 6 = ox # of S +6 +4 +5 +5 +3 +4

  6. Redox Reactions To see if a redox reaction has occurred, check to see if the oxidation numbers of any element involved in the reaction have changed. Mg(s) + Fe(NO3)2 (aq) Mg(NO3)2(aq)+ Fe (s) ox #0 +2 +5 -2 +2 +5 -2 0 • Mg goes from an element to an ion: Mg  Mg2+ • Mg got oxidized. Fe2+ was the oxidizing agent. • Fe goes from an ion to an element: Fe2+  Fe • Fe2+ got reduced. Mg was the reducing agent. • This type of redox reaction, where the ion in solution is replaced through the oxidation of an element, is called a single replacement reaction. Oxidation Is Loss of electrons Reduction Is Gain of electrons

  7. Redox Reactions • Single replacement reactions have the form: • A + BX (aq)  AX (aq) + B • Single replacement reactions will occur if the element A is more likely to get oxidized than the ion B. • The activity series gives this information. See the text. Oxidation Is Loss of electrons Reduction Is Gain of electrons • Notice that H2 (g) is included in this series. The series can tell us whether a metal will dissolve in a nonoxidizing acid.

  8. Will the following pairs react? Zn + Cu(NO3)2 (aq) Cu + Zn(NO3)2 (aq) Zn + NaNO3 (aq) Au + AgNO3 (aq) Ag + Au(NO3)3 (aq) Will the following metals dissolve in acid? Li Al Ni Cu Au What gas will be produced? H2 y n n n y y y y n n

  9. Redox Reactions - Example Molecular equation: Mg(s) + Fe(NO3)2 (aq) Mg(NO3)2(aq)+ Fe (s) Total Ionic equation: Mg(s) + Fe2+(aq) + 2NO3-(aq) Mg2+(aq)+ Fe(s) + 2NO3-(aq) Net Ionic equation: Mg(s) + Fe2+ (aq) Mg2+(aq)+ Fe (s) Mg got oxidized. Fe2+ was the oxidizing agent. Fe2+ got reduced. Mg was the reducing agent.

  10. Redox Reactions - Example When Zn dissolves in sulfuric acid, the molecular equation is Zn(s) + H2SO4 (aq) ZnSO4(aq)+ H2 (g), the total ionic equation is Zn(s) + 2H+(aq) + SO42-(aq) Zn2+(aq) + SO42-(aq) + H2 (g), and the net ionic equation is: Zn(s) + 2H+(aq) Zn2+(aq) + H2 (g) Zn got oxidized. H+ was the oxidizing agent. H+ got reduced. Zn was the reducing agent.

  11. Balancing Redox Reactions • A redox reaction consists of an oxidation AND a reduction. • Although an oxidation cannot occur without a reduction, it is useful to be able to look at each separately. • A redox reaction can be considered to be the sum of an oxidation half reaction and a reduction half reaction.

  12. Balancing Redox Reactions The Method of Half-Reactions: Step 1. Use oxidation numbers to identify which species is oxidized and which is reduced. Cr2O72- (aq) + Fe2+(aq)  Cr3+ (aq) + Fe3+(aq) +6 -2 +2 +3 +3 Cr2O72- is reduced. Fe2+ is oxidized.

  13. The Method of Half-Reactions Step 2. Separate the reaction into the two half-reactions. oxidation half-reaction: Fe2+(aq)  Fe3+(aq) reduction half-reaction: Cr2O72- (aq)  Cr3+ (aq)

  14. The Method of Half-Reactions Step 3. Balance each half-reaction. Step 3a. Balance the elements other than H and O Fe2+(aq)  Fe3+(aq)already balanced Cr2O72- (aq)  Cr3+ (aq) Cr2O72- (aq) 2Cr3+ (aq) Step 3b. Balance O by adding H2O: Fe2+(aq)  Fe3+(aq)already balanced Cr2O72- (aq)  2Cr3+ (aq) Cr2O72- (aq)  2Cr3+ (aq)+ 7H2O(l)

  15. The Method of Half-Reactions Step 3c. Balance H by adding H+: Fe2+(aq)  Fe3+(aq)already balanced Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l) 14H+ (aq) +Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l) Step 3d. Balance the charge by adding electrons (e-) to the side that is relatively more positive: Fe2+(aq)  Fe3+(aq)+ e- 6e- +14H+(aq) + Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l)

  16. The Method of Half-Reactions Step 4. Multiply each half-reaction by an integer so that the electrons (e-) in each equation balance: 6(Fe2+(aq)  Fe3+(aq)+ e-) 6e- +14H+ (aq) + Cr2O72- (aq)  2Cr3+ (aq) + 7H2O(l) Step 5. Add the two half-reactions, canceling species where possible: 6Fe2+(aq)+6e-+14H+(aq)+Cr2O72-(aq)6Fe3+(aq)+6e-+2Cr3+(aq)+7H2O(l) 6Fe2+(aq) + 14H+(aq)+ Cr2O72-(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) this is the balanced equation

  17. The Method of Half-Reactions Step 6. CHECK THE EQUATION to make sure it is balanced both for mass and for charge. 6Fe2+(aq) + 14H+(aq)+ Cr2O72-(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

  18. The Method of Half-Reactions Reactions in Basic solution If the redox reaction takes place in BASIC solution, use steps 1-6 (as before) to balance the equation as if it took place in acidic solution. Then perform one more step: Step 7. (ONLY for redox reactions taking place in basic solution!) Add OH- to BOTH sides of the equation to cancel all of the H+, then make sure H2O appears only on one side of the equation. If, in our example, the reaction between Fe3+ and Cr2O72- took place in basic solution: 6Fe2+(aq) + 14H+(aq) + 14OH- (aq)+ Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH-(aq)

  19. The Method of Half-Reactions 6Fe2+(aq) + 14H+(aq) + 14OH- (aq)+ Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq) Remember that H+(aq) + OH –(aq)  H2O(l): 14H+(aq) +14OH- (aq)  14H2O(l) 6Fe2+(aq) + 14H2O(l)+ Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq)

  20. The Method of Half-Reactions Reactions in Basic solution Step 7. Add OH- to BOTH sides of the equation to cancel all of the H+, then make sure H2O appears only on one side of the equation. 7H2O 6Fe2+(aq) + 14H2O(l) + Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 14OH- (aq) Balanced equation for the reaction in basic solution: 6Fe2+(aq) + 7H2O(l) + Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 14OH- (aq)

  21. Stoichiometry of Redox Reactions Once you have a balanced redox equation, you may use it to perform stoichiometric calculations. Example: How many grams of potassium dichromate are needed to oxidize 10.0 g of iron(II) nitrate according to the following balanced equation? 6Fe2+(aq) + 7H2O(l) + Cr2O72-(aq) 6Fe3+(aq) + 2Cr3+(aq) + 14OH- (aq) 10.0 g Fe(NO3)2 x 1 mol Fe(NO3)2 x 1 mol Fe2+ x 1 mol Cr2O72- 179.86 g Fe(NO3)2 1 mol Fe(NO3)2 6 mol Fe2+ x 1 mol K2Cr2O7 x 294.18 g K2Cr2O7 = 2.73 g K2Cr2O7 1 mol Cr2O72- 1 mol K2Cr2O7

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