well i don't really know the word chemical equation, but the balanced form is:
2KI(aq) + Br(aq) --> I2(s) + 2KBr(aq)
so if you just figure out the word form of the elements in the equation you'll figure it out..
Bromine and Potassium iodide react to form Potassium bromide and Iodine.
Yes, liquid bromine can react with potassium iodide to form potassium bromide and elemental iodine. This reaction is a displacement reaction, where the more reactive element (bromine) displaces the less reactive element (iodine).
The reaction between potassium iodide and bromine produces potassium bromide and iodine. This is a redox reaction where bromine gets reduced to bromide ions, while iodide ions get oxidized to form elemental iodine. The balanced chemical equation for this reaction is 2 KI + Br2 → 2 KBr + I2.
Br2 (g) + 2 KI (aq) 2KBr (aq) + I2 (s)
Br2 + (2e)- --> 2 Br- 2I- --> I2 + (2e)-
Bromine and Potassium iodide react to form Potassium bromide and Iodine.
Yes, liquid bromine can react with potassium iodide to form potassium bromide and elemental iodine. This reaction is a displacement reaction, where the more reactive element (bromine) displaces the less reactive element (iodine).
In a reaction between bromine and potassium iodide, bromine acts as an oxidizing agent, not a reducing agent. It oxidizes iodide ions to form iodine molecules while itself being reduced to bromide ions.
When potassium iodide is added to a solution of bromine, a reaction occurs forming potassium bromide and iodine. The iodine produced in the reaction is responsible for the color change in the solution from orange to brown/black.
The reaction between potassium iodide and bromine produces potassium bromide and iodine. This is a redox reaction where bromine gets reduced to bromide ions, while iodide ions get oxidized to form elemental iodine. The balanced chemical equation for this reaction is 2 KI + Br2 → 2 KBr + I2.
potassium
Br2 (g) + 2 KI (aq) 2KBr (aq) + I2 (s)
The reaction between bromine water and potassium iodide forms potassium bromide and iodine. This is a redox reaction where bromine is reduced to bromide ions and iodide ions are oxidized to form iodine. The solution will change color from orange/brown to colorless as the iodine precipitates out.
Br2 + (2e)- --> 2 Br- 2I- --> I2 + (2e)-
This is a double displacement reaction where bromine (Br2) reacts with potassium iodide (KI) to form potassium bromide (KBr) and iodine (I2) by exchanging ions. The bromine displaces the iodine from potassium iodide to form potassium bromide and free iodine.
The reaction between bromine and aqueous strontium iodide is a double displacement reaction. Bromine replaces iodide in strontium iodide, forming strontium bromide and iodine. The balanced chemical equation is: Br2 + SrI2 --> SrBr2 + I2.
Its actually: 2KI(aq)+Br2(aq)-> I2(s)+2KBr(aq)