HalfReactionBalancing

3 in problem 1a the half reactions show clo4 gaining

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Unformatted text preview: ─ + 4H+ + NO3─ S + 2 e─ NO + 2 H2O Get the electrons to be the same on both sides, and then add the two half-reactions. 3 S + 6 e─ (3x) 3 S2─ 6 e─ + 8 H+ + 2 NO3─ 3 S2─ + 8 H+ + 2 NO3─ 2NO + 4 H2O (2x) 3 S + 2 NO + 4 H2O Step 4: 3 S, 8H, 2 N, 6 O atoms on both sides; zero net charge on both sides. Check! 2. In problem 1a, the half-reactions show Br─ giving away electrons: acting as a reducing agent. In problem 1b, the half-reactions show Ni giving away electrons: acting as a reducing agent. Metal atoms often act as reducing agents. In problem 1c, the half reactions show the reactant S2─ donating electrons, which is the behavior of a reducing agent. 3. In problem 1a, the half-reactions show ClO4─ gaining electrons, since the Cl atom is changing to a more negative oxidation number in the reaction. Gaining electrons means ClO ─ is being reduced. 4 In problem 1b, the half-reactions show the N atom in the NO3─ is having electrons added to it in the reaction: the NO3─ is being reduced. In problem 1c, the half-reactions show the N atom in the NO3─ is again having electrons added to it in the reaction: the NO3─ is being reduced. ***** ©2009 ChemReview.net v. k1 Page 392 Module 16: Half-Reaction Balancing Lesson 16D: Balancing Redox With Spectator Ions Present Balancing oxidation-reduction reactions can be tricky when spectator ions (ions that do not change in the reaction) are included in the equation. Consider this redox reaction carried out in an aqueous solution. K2Cr2O7 + HNO3 + C2H6O KNO3 + C2H4O + H2O + Cr(NO3)3 Balancing this equation by trial and error could take some time. Balancing using oxidation numbers can help, but for complex reactions, using half-reactions to balance is usually faster. But when the spectators are present, how do you find the half-reactions? The following steps present a system to break complex redox reactions into half-reactions. In your notebook, apply the following steps to the reaction above. Steps for Balancing Redox With Spectators Present 1. The fundamental rule: To understand the reactions of ionic compounds, re-write the reaction using separated-ion formulas. You may leave out coefficients at this step. The rules for identifying ionic compounds (Lesson 7A), separating ionic solid into separated-ion formulas (Lesson 7C), and strong acid ionization (Lesson 14A) include: a. Compounds with both metal and non-metal atoms are usually ionic. b. Compounds containing alkali metal atoms (Li, Na, K, Rb, Cs, Fr) are soluble in water and separate ~100% to form monatomic +1 ions. c. Nitrates dissolve and ionize ~100% in water to form nitrate ions (NO3―). d. Aqueous solutions of strong acids (such as HCl and HNO3) ionize ~100%. Try Step 1 on the above reaction, then check your answer below. ***** The separated-ions version of the equation, leaving out coefficients, is K+ + Cr2O72― + H+ + NO3― + C2H6O K+ + NO3― + C2H4O + H2O + Cr3+ + NO3― 2. Cross out the spectators: particles that do not change in the react...
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