# 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

80% (102 ratings) ###### Problem Details

solve

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.