Problem: solveCl2(g) + 2LiBr(aq) → 2LiCl(aq) + Br2(l)a) Cl (in Cl2) is oxidized; Cl2 is the reducing agent. Br− (in LiBr) is reduced; LiBr is the oxidizing agent.b) Br− (in LiBr) is oxidized; LiBr is the oxidizing agent. Cl (in Cl2) is reduced; Cl2 is the reducing agent.c) Br− (in LiBr) is oxidized; LiBr is the reducing agent. Cl (in Cl2) is reduced; Cl2 is the oxidizing agent.d) Cl (in Cl2) is oxidized; Cl2 is the oxidizing agent. Br− (in LiBr) is reduced; LiBr is the reducing agent.

FREE Expert Solution

Cl2 (g)  elemental  O.S. = 0


LiBr

O.S. Li + O.S. Br = 0

(+1) + O.S. Br = 0

O.S. Br = -1


LiCl

O.S. Li + O.S. Cl = 0

(+1) + O.S. Cl = 0

O.S. Cl = -1


Br2 (g)  elemental  O.S. = 0


Change in O.S.

Li: +1 to +1 no change

Cl: 0 to -1  decrease/gain electrons  reduction  oxidizing agent

Br: -1 to 0 increase/lose electrons  oxidation reducing agent


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Problem Details

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Cl2(g) + 2LiBr(aq) → 2LiCl(aq) + Br2(l)

a) Cl (in Cl2) is oxidized; Cl2 is the reducing agent. Br (in LiBr) is reduced; LiBr is the oxidizing agent.

b) Br (in LiBr) is oxidized; LiBr is the oxidizing agent. Cl (in Cl2) is reduced; Cl2 is the reducing agent.

c) Br− (in LiBr) is oxidized; LiBr is the reducing agent. Cl (in Cl2) is reduced; Cl2 is the oxidizing agent.

d) Cl (in Cl2) is oxidized; Cl2 is the oxidizing agent. Br (in LiBr) is reduced; LiBr is the reducing agent.

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Our tutors have indicated that to solve this problem you will need to apply the Calculate Oxidation Number concept. If you need more Calculate Oxidation Number practice, you can also practice Calculate Oxidation Number practice problems.