Question

use the method of half reactions to balance the following equations: a) CrI3 (s) + Cl2 (g) ----> CrO 2-4 (aq) + IO-4 (aq) (basic) b) C2H5OH (aq) + I-3 (aq) ----> CO2 (g) + CHO-2 (aq) + CHI3 (aq) + I- (aq) (acidic)

Answer 1

a) CrI3(s) + Cl2(g) ---> CrO2^4-(aq) + IO4^-(aq) + Cl-(aq)

first half reaction,

CrI3 ---> CrO4^2- + IO4^-

balance I,

CrI3 ---> CrO4^2- + 3IO4^-

balance O,

CrI3 + 16H2O ---> CrO4^2- + 3IO4^-

balance H,

CrI3 + 16H2O ---> CrO4^2- + 3IO4^- + 32H+

add OH- for basic medium,

CrI3 + 16H2O + 32OH- ----> CrO4^2- + 3IO4^- + 32H2O

add e-.

CrI3 + 16H2O + 32OH- ---> CrO4^2- + 3IO4^- + 32H2O + 27e-

second half reaction,

Cl2 ---> Cl-

balance Cl, and add e-

Cl2 + 2e- ---> 2Cl-

multiply above reaction with 27 and the first half reaction with 2 and add them together,

2CrI3 + 32H2O + 64OH- ---> 2CrO4^2- + 6IO4^- + 64H2O + 54e-

27Cl2 + 54e- ---> 54Cl-

----------------------------------------------------------------------------------------------------------------------

2CrI3(s) + 27Cl2(g) + 64OH-(aq) ----> 2CrO4^2-(aq) + 6IO4^-(aq) + 54Cl-(aq) + 32H2O(l)

Is the balanced equation

b) C2H5OH(aq) + I3^-(aq) ----> CO2(g) + CHO2^-1(aq) + CHI3(aq) + I-(aq)

first half reaction,

C2H5OH ---> CHO2^- + CO2

balance O

C2H5OH + 3H2O ---> CHO2^- + CO2

balance H,

C2H5OH + 3H2O ---> CHO2^- + CO2 + 11H+

add e-,

C2H5OH + 3H2O ---> CHO2^- + CO2 + 11H+ + 10e-

second half reaction,

I3^- ---> I-

balance I and add e-,

I3 + 2e- ----> 3I-

Add both half reaction by adjusting e-'s,

2C2H5OH + 6H2O ---> 2CHO2^- + 2CO2 + 22H+ + 20e-

10I3 + 20e- ----> 30I-

---------------------------------------------------------------------------------------

C2H5OH + 5I3^- + 3H2O ---> CHO2^- + 15I- + CO2 + 11H+

Is the net balance equation