Balance the following half-reactions by adding the appropriate number of electro

Question

Balance the following half-reactions by adding the appropriate number of electrons. Which are oxidation half-reactions and which are reduction half-reactions? Use the left and right arrow keys to move the cursor out of a superscript or subscript in the module. Add electrons by typing e^- with the appropriate coefficient. For example, to add 2 electrons, type 2e^- on the appropriate side of the equation. Fe^2+ (aq) right arrow Fe^+3 (aq) Oxidation Reduction Agl(s) right arrow Ag(s) +I^-(aq) Oxidation Reduction VO^+2 (aq) + 2H^+ (aq) right arrow VO^2+ + H2O(l) Oxidation Reduction I2(s)+ 6H2O(l) right arrow 2IO3^- (aq)+12 H^+(aq) Oxidation Reduction

Explanation / Answer

a. Fe2+(aq) ---> Fe3+(aq)

Fe is going from +2 to +3, and looses 1 e- in the process so, it is an oxidation reaction.

Balanced eqation : Fe2+(aq) ---> Fe3+(aq) + 1e^-

b. AgI(s) ---> Ag(s) + I-(aq)

Ag is getting reduced from +1 to 0 state by gaining an e-, so it is an reduction reaction.

Balanced equation : AgI(s) + 1e^- ---> Ag(s) + I-(aq)

c. VO2^+(aq) + 2H+(aq) ---> VO^2+ + H2O(l)

V is going from +5 to +4 state by gaining 1e-, so it an reduction reaction

Balanced equation : VO2^+(aq) + 2H+(aq) + 1e^- ---> VO^2+(aq) + H2O(l)

d. I2(s) + 6H2O(l) --> 2IO3-(aq) + 12H+(aq)

I goes from 0 to +5 state, so it is an oxidation reaction.

Balanced equation : I2(s) + 6H2O(l) ---> 2IO3-(aq) + 12H+(aq) + 5e-

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