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an example for redox reaction????? |
Example 1 Balance the equation from the following two half-reactions: Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O H2C2O4 --> 2 CO2 + 2 H+ + 2 e- Solution Multiply the second equation by 3 and then add them algebraically so that the electrons in the two half-reaction equations cancel completely. Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O 3 H2C2O4 --> 6 CO2 + 6 H+ + 6 e- add the two equations and cancel the electrons to give the over all equation Cr2O72- + 8 H+ + 3 H2C2O4 --> 2 Cr3+ + 7 H2O + 6 CO2 Discussion This example illustrate balancing redox equations in acid solutions. Example 2 Balance the equation from the two half-reactions: Cd --> Cd2+ + 2 e- 4 H+ + NO3- + 3 e- --> NO + 2 H2O Solution The first half-reaction has 2 electrons, whereas the second one has 3. The lowest common multiple of 2 and 3 is 6. Thus, you multiply the first equation by 3 and the second one by 2. The half-reaction equations become 3 Cd --> 3 Cd2+ + 6 e- 8 H+ + 2 NO3- + 6 e- --> 2 NO + 4 H2O add the two equations and cancel the electrons to give the overvall equation 3 Cd + 8 H+ + 2 NO3- --> 3 Cd2+ + 2 NO + 4 H2O Discussion The last step in balancing oxidation and reduction reactions is simple. Example 3 In a basic solution, Fe(OH)2 and Fe(OH)3 are solids. The former may be oxidized by H2O2. Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O. Balance this equation. Solution The balanced half-reactions are: Fe(OH)2 + OH- --> Fe(OH)3 (s) + e- H2O2 + 2 e- --> 2 OH- Thus, the balanced equation is 2 Fe(OH)2 + H2O2 --> 2 Fe(OH)3 Discussion Note that the balanced equation does not have an H2O in it. Example 4 Balance the following reaction, which is carried out in an acidic solution: I- + IO3- --> I2 Solution The half-reactions are: 2 I- --> I2 + 2 e- (oxidized) 2 IO3- + 10 e- --> I2 (reduced) The balanced equation is 6 H+ + 5 I- + IO3- ® 3 I2 + 3H2O Discussion You may study Electro chemistry to gain better insight to the oxidation reduction process. |