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Activity Series. We are going to ignore the second half for today And we will ignore the halogens for now Leaving a list of metals. This list has the most reactive metals on the top and the least reactive metals on the bottom.

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Activity Series


  • We are going to ignore the second half for today

  • And we will ignore the halogens for now

  • Leaving a list of metals


  • This list has the most reactive metals on the top and the least reactive metals on the bottom.

  • Meaning the elements in the top would rather be in a compound more then the elements on the bottom.


  • A single replacement reaction will only happen if the element by itself is higher on the list then the element in the compound.

  • For example:

  • Ni + KOH 

  • Find Ni

  • Find K

  • Ni is lower then K so this will not react. We would write “No rxn”

No rxn


Cu + AlCl3

  • Here is another example:

  • Al + CuCl2

  • Find Al

  • Find Cu

  • Al is higher then Cu so this can react.

  • We can then write the products.

  • What element will be by itself now?

  • What will the compound look like?

  • Balance


Ag + Mn(NO3)2

  • Here is another example:

  • Mn + AgNO3

  • Find Mn

  • Find Ag

  • Mn is higher then Ag so this can react.

  • We can then write the products.

  • What element will be by itself now?

  • What will the compound look like?

  • Mn will have a +2 charge

  • In other problems it will be Mn(II)

  • Balance


No rxn

  • Here is another example:

  • Sn(IV) + NaCN

  • Find Sn

  • Find Na

  • Sn is lower then Na so this is no reaction


Practice

  • Na + ZnSO4

  • Cr(II) + Fe2O3 

  • Pb(II) + CaCO3 

  • Fe(III) + PbI2 

  • K + LiOH 


The halogen

  • This works the same way with the halogens

  • Example:

  • Br2 + NaCl

  • Bromine is below chlorine so its no reaction.


The halogen

  • Example:

  • F2 + KI 

  • Fluorine is above Iodine so this will react.

  • What element will be by itself now?

  • What will the compound look like?

  • Balance


Practice

  • F2 + AlCl3

  • I2 + ZnCl2 

  • Cl2 + FeBr2 


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