Br2 + (2e)- --> 2 Br-
2I- --> I2 + (2e)-
Bromine and Potassium iodide react to form Potassium bromide and Iodine.
Pb2+ + I- --> PbI2(s)potassium and acetate ions are left out of the equation, because they don't react (stay unchanged in solution)
2KI + F2 ----> 2KF + I2
Yes, it can, by displacing the Iodide
Its actually: 2KI(aq)+Br2(aq)-> I2(s)+2KBr(aq)
potassium
Bromine and Potassium iodide react to form Potassium bromide and Iodine.
A reaction doesn't occur.
Pb2+ + I- --> PbI2(s)potassium and acetate ions are left out of the equation, because they don't react (stay unchanged in solution)
2KI + F2 ----> 2KF + I2
Cl2 + 2KI --> 2KCl + I2
Yes, it can, by displacing the Iodide
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
H2O2 + 2 KI --> 2 KOH + I2
Its actually: 2KI(aq)+Br2(aq)-> I2(s)+2KBr(aq)
you have to write... 2KI + Cl2 = 2KCl + I2
2KI + Cl2 = 2KCl + I2