o2SbO+ + 2Cd2++ 3H2OSb2O5 + 2Cd+ 6H+ In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

Question

2SbO+ + 2Cd2++ 3H2OSb2O5 + 2Cd+ 6H+

In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

1. 
   

name of the element oxidized:			name of the element reduced:
formula of the oxidizing agent:		formula of the reducing agent:

Answer 1

Sb in SbO^+ is 3+

Sb in Sb2O5 is 5+

Sb has been oxidized

Cd^2+ On left side is 2+

Cd as Cd on right side is zero

Cd^2+ has been reduced

Cd^2+ is the oxidizing agent

SbO^+ is the reducing agent

Answer 2

Cd goes from +2 to 0. It is gaining electrons and being reduced.

Sb is going from 3+ to 5+. It is losing electrons and being oxidized.

O stays at -2 throughout; H stays at +1 throughout.