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21.31 Balance each ske whether the reaction is (a) Co(s) + H* (aq) (b) Mn2 (aq) + Br,(1) (c) Hg,2*(aq) Hg ->

21.31 Balance each ske whether the reaction is (a) Co(s) + H* (aq) (b) Mn2 (aq) + Br,(1) (c) Hg,2*(aq) Hg ->

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Please show all solutions. And please balnce using half reaction method following the the steps shown in the second picture. Thank you.

Steps in the Half-Reaction Method The balancing process begins with a "skel- eton" ionic reaction that consists of only species that are oxidized and reduced. Here are the steps in the half-reaction method: Step 1. Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of the species: if the oxidized form of a species is on the left side, the reduced form must be on the right, and vice versa. Step 2. Balance the atoms and charges in each half-reaction. Atoms are balanced in this order: atoms other than O and H, then O, then H. Charge is balanced by adding electrons (e-) to the left side in the reduction half-reaction because the reactant gains them and to the right side in the oxidation half-reaction because the reactant loses them. Step 3. If necessary, multiply one or both half-reactions by an integer so that number of e gained in reduction number of e- lost in oxidation %3D Step 4. Add the balanced half-reactions, and include states of matter. Step 5. Check that the atoms and charges are balanced. Let's balance a redox reaction that occurs in acidic solution and then see how to bal- ance one in basic solution in Sample Problem 21.1.
Transcribed Image Text:Steps in the Half-Reaction Method The balancing process begins with a "skel- eton" ionic reaction that consists of only species that are oxidized and reduced. Here are the steps in the half-reaction method: Step 1. Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of the species: if the oxidized form of a species is on the left side, the reduced form must be on the right, and vice versa. Step 2. Balance the atoms and charges in each half-reaction. Atoms are balanced in this order: atoms other than O and H, then O, then H. Charge is balanced by adding electrons (e-) to the left side in the reduction half-reaction because the reactant gains them and to the right side in the oxidation half-reaction because the reactant loses them. Step 3. If necessary, multiply one or both half-reactions by an integer so that number of e gained in reduction number of e- lost in oxidation %3D Step 4. Add the balanced half-reactions, and include states of matter. Step 5. Check that the atoms and charges are balanced. Let's balance a redox reaction that occurs in acidic solution and then see how to bal- ance one in basic solution in Sample Problem 21.1.
21.31 Balance each skeleton reaction, calculate El and state whether the reaction is spontaneous: (a) Co(s) + H* (aq) (b) Mn2 (aq) + Br,(1) (c) Hg,2 (aq) cell Co2+(aq) + H,(g) MnO, (aq) + Br (aq) [acidic] -> Hg2 (aq) + Hg(1)
Transcribed Image Text:21.31 Balance each skeleton reaction, calculate El and state whether the reaction is spontaneous: (a) Co(s) + H* (aq) (b) Mn2 (aq) + Br,(1) (c) Hg,2 (aq) cell Co2+(aq) + H,(g) MnO, (aq) + Br (aq) [acidic] -> Hg2 (aq) + Hg(1)
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