Cl_2 (g) + Cr(OH)_3 (s) -> Cl^- (aq) + CrO4^2- (aq) __________ 1. The oxidation
ID: 689988 • Letter: C
Question
Cl_2 (g) + Cr(OH)_3 (s) -> Cl^- (aq) + CrO4^2- (aq)__________ 1. The oxidation number of Cr in CrO4^2- is (1)_.
__________ 2. There are (2)_ electrons lost in the balancedoxidation half reaction.
__________ 3. There are (3)_ electrons gained in the balancedreduction half reaction.
__________ 4. There are (4)_ moles of water produced in the overallequation.
__________ 5. In the overall equation, the coefficient for Cl^- is(5)_.
__________ 6. The reducing agent is (6)_.
Could someone show all the steps involved, and how to do eachnumber.
I vaguely remember how to solve redox reactions..
Explanation / Answer
Cl_2 (g) + Cr(OH)_3 (s) -> Cl^- (aq) + CrO4^2- (aq) Cr is in +6 state. get it by solving: x + 4*(-2) = +2 No there are 3 electrons lost. Write the balanced half reaction tofind it out· yes there are 3 electrons gained in the balanced reduction halfreaction. There are 4 moles of water produced in the overallequation. In the overall equation, the coefficient for Cl^- is 2 The reducing agent is Cr(OH)_3