Electrochemical Series ( ECS )
Cations
(MALE)
|
Acronym
|
Acronym
|
Anions
(FEMALE)
|
K+
|
Kalau
|
Fly
|
F-
| |||||
Na+
|
Nak
|
Sampai
|
SO42-
| |||||
Ca2+
|
Kahwin
|
Nak
|
NO3-
| |||||
Mg2+
|
Mesti
|
Kantoi
|
Cl-
| |||||
Al
|
Ada
|
Baru
|
Br-
| |||||
Zn2+
|
Zakat
|
Insaf
|
I-
| |||||
Fe2+
|
Fitrah
|
Oh
|
OH-
| |||||
Sb2+
|
Supaya
| |||||||
Pb2+
|
Pasangan
| |||||||
Hg
|
Hidup
| |||||||
Cu2+
|
Cukup
| |||||||
Ag+
|
Anak
|
KING AND SLAVE RULE
ECS frequently used to solve neutralization problem and so on.
Lets consider the highest elements / ions in ECS are Kings and Queen whereas the lowest part are slaves.
We know that Kings and Queens will not marry with the slaves.
Symbol for "cross-form-match" is [ X ]
Symbol for "equal-form-match" is [ = ]
Symbol for "cross-form-match" is [ X ]
Symbol for "equal-form-match" is [ = ]
Problem 1:
A beaker filled with 50 ml of 1M Sodium Hydroxide solution and 50 ml of 1M Zinc Sulphate solution. Will the reaction occur in the beaker?
First step to do is listing all the initial compound and ions exist in the beaker.
NaOH -----> Na + + OH -
ZnSO4 -----> Zn 2 + + SO42-
According to ECS, Na placed higher than Zn in the cations group, so Na will be assumed as the King and Zn as a slave.
This rule also applied in the anions group where SO4 placed higher than OH anion, then SO4 is assumed to be Queen and OH will be slaves.
Initial solution shows that the King Na married with the slave OH,and the Queen SO4 is married with slave Zn.
The king should marry with queen as we known, then the reaction will occur in the beaker to match the compound correctly based on this rule.
When there is a "cross-form-match" compound in the initial state, the reaction will occur to match the compound until it form the "equal-form-match" compound.
This is a "cross-form-match" compound.
This is an "equal-form match" compound.
Problem 2:
A beaker filled with 50 ml of 1M Potassium Flouride solution and 50 ml of 1M Copper Nitrate solution. Will the reaction occur in the beaker?
First step to do is listing all the initial compound and ions exist in the beaker.
KF ---> K+ + F-
Cu(NO3)2 ---> Cu2+ + 2 ( NO3- )
K cations is placed higher than Cu cations. It makes K is the king and Cu is the slaves.
F anions is higher than NO3 anions. It makes F is the Queen and NO3 is the slaves.
When there is a "cross-form-match" compound in the initial state, the reaction will occur to match the compound until it form the "equal-form-match" compound.
Problem 2:
A beaker filled with 50 ml of 1M Potassium Flouride solution and 50 ml of 1M Copper Nitrate solution. Will the reaction occur in the beaker?
First step to do is listing all the initial compound and ions exist in the beaker.
KF ---> K+ + F-
Cu(NO3)2 ---> Cu2+ + 2 ( NO3- )
K cations is placed higher than Cu cations. It makes K is the king and Cu is the slaves.
F anions is higher than NO3 anions. It makes F is the Queen and NO3 is the slaves.
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