Balance the following half-reactions by adding the appropriate number of electro

Question

Balance the following half-reactions by adding the appropriate number of electrons. Which are oxidation half-reactions and which are reduction haIf-reactions? Use the left and right arrow keys to move the cursor out of a superscript or subscript in the module, Add electrons by typing e^- with the appropriate coefficient, For example, to add 2 electrons, type 2e^- on the appropriate side of the equation. Fe2+ (aq) rightarrow Fe3+ (aq) Oxidation Reduction AgI(s) rightarrow Ag(s) + I-(aq) Oxidation Reduction VO2+ (aq) + 2H+ (aq) rightarrow VO2+ + H2O(l) Oxidation Reduction I2 (s) + 6H2O(l) rightarrow 2IO3-(aq) + 12H+ (aq) Oxidation Reduction

Explanation / Answer

I recommend balancing the reaction in acid solution first, then converting to a basic solution. Here are the steps for balancing in an acid solution (adding H+ and H2O). Separate the reaction into two half-reactions, one for Zn and one for N in this case. Zn(s) -----> Zn(OH)42- (aq) NO31- -----> NH3 Check to make sure the main atoms, Zn and N are balanced. They already are in this case. Balance the O by adding water as needed. Zn(s) 4H2O -----> Zn(OH)42- NO31- -----> NH3 + 3H2O Next, balance the H by adding H+ has needed. Zn(s) 4H2O -----> Zn(OH)42- + 4H+ NO31- + 9H+ -----> NH3 + 3H2O Balance the charges by adding electrons (e-) to the more positive side of each half-reaction. Zn(s) 4H2O -----> Zn(OH)42- + 4H+ + 2e- NO31- + 9H+ + 8e- -----> NH3 + 3H2O Make the electrons lost and gained equal, in this case by multiplying the first half-reaction by 4. 4Zn(s) 16H2O -----> 4Zn(OH)42- + 16H+ + 8e- NO31- + 9H+ + 8e- -----> NH3 + 3H2O Add the two half-reactions together. Simplify H2O and H+ where possible. 4Zn(s) 16H2O -----> 4Zn(OH)42- + 16H+ + 8e- NO31- + 9H+ + 8e- -----> NH3 + 3H2O 4Zn (s) + 13H2O + NO31- -----> 4Zn(OH)42- + 7H+ + NH3 (after adding) Notice that both sides of the equation have a -1 charge and that all atoms are balanced. The equation is balanced for an acidic (H+, H2O) solution. To balance for a basic solution, add the same number of OH- to each side of the equation as there are H+. 4Zn (s) + 13H2O + NO31- + 7OH- -----> 4Zn(OH)42- + 7H+ + 7OH- + NH3 Combine the H+ and OH- to make H2O. 4Zn (s) + 13H2O + NO31- + 7OH- -----> 4Zn(OH)42- + 7H2O + NH3 Simplify the H2O. 4Zn (s) + 6H2O + NO31- + 7OH- -----> 4Zn(OH)42- + NH3 The equation is now balanced for a basic (OH-, H2O) solution. Both sides have a net charge of -8 and all the atoms are balanced.

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