In each of the following balanced oxidation-reduction equations, identify those

Question

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case.

a. I2O3 (s) + 5 CO --> I2 (s) + 5 CO2 (g)

b. 2Hg +2 (aq) + N2H4 (aq) --> 2 Hg (l) + N2 (g) + 4 H+ (aq)

c. 3 H2S (aq) + 2 H+ (aq) + 2 NO3- (aq) --> 3 S (s) + 2 NO (g) + 4 H2O (l)

d. Ba +2 (aq) + 2 OH- (aq) + H2O2 (aq) + 2 ClO2 (aq) --> Ba (ClO2)2 + 2 H2O(l) + O2 (g)

Please show step by step, I'm so lost. Thank you :)

Explanation / Answer

m = oxidation number

a) I ( -1 to 0) , O ( +2/3 to -2) , C (+2 to +4)

in I2O3 m of I is -1, so that of O is 2 x (-1) + 3 mo = 0 ==> m of O is +2/3
in CO m of O is -2, so that of C is +2
in I2 m of I is zero.
in CO2 m of O is -2 so that of C is +4.

b)Hg(+2 to 0) , N (-2 to 0)

Here, Hg+2 ion goes to liquid Hg state.(nuetral)

in N2H4 m of H is +1 so that of N is -2
in N2 m is zero.

c) S ( -2 to 0) , N (+5 to +2)

in H2S m of H is +1 and that of S is -2
in N03- m of O is -2 so that of N is -(3 x (-2)) -1 = +5

in S its zero
in NO m of O is -2 , so m of N is +2

d) O ( -2 , -1 , +1/2 to 0 , -2)

in OH- m of O is -2
in H2O2 , m of H is +1 so 2 x 1 + 2 x m = 0 ==> m of O = -1

in ClO2 m of Cl is -1 so that of O is +1/2

in Ba(ClO2)2 ==> ClO2- m of Cl is -1 so that of O is zero.
in H2O m of O is -2
in O2 m of O is 0...

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