In the electrochemistry lab we assumed that the electrode process at the copper
ID: 1020095 • Letter: I
Question
In the electrochemistry lab we assumed that the electrode process at the copper cathode in the sulfuric acid solution was 100% efficient and no other reactions were taking place at this electrode cathode. What if water was reduced at the copper-wire cathode as well as the hydrogen ions?
2H20 + 2e- ----> H2(g) + 2OH- (aq)
What affect would this have on the calculated value of the equivalent mass for the unknown metal (increase, decrease, or remain the same). Explain/give reasoning for your answer.
*Not sure if this may help below*
Ered = -0.83 // 2H2O (l) + 2e- ----> H2(g) + 2OH- (aq)
Ered = 0.34 // Cu +2 (aq) + 2e- ----> Cu (s)
Explanation / Answer
We consider that in electrode reaction with copper cathode and sulfuric acid that the reaction is 100% complete and no other reaction takes place.
But, let us consider a reaction in which water is reduced to copper cathode as well as hydrogen ion as below:
2H2O (l) + 2e- ----> H2(g) + 2OH- (aq)
Now, here we can see the generation of OH- (aq) ion. So, it will certainly react with Cu +2 (aq) ion. Now, the reaction will go as below:
Cu +2 (aq) + 2OH- (aq)----> Cu(OH)2 (s)
So, here we can see there will be some generation of copper (II) hydroxide precipitate (blue colored precipitate). So, this will be an 'impurity' which will be deposited over the copper wire.
Now, from the reaction equation we can see that the mole ratio of Cu +2 (aq) and Cu(OH)2 (s) is 1:1
Now,
1 mole of Cu metal is equivalent to 63.55 g
1 mole of Cu(OH)2 is equivalent to 97.56 g (molar mass)
So, the mass of 1 mole of Cu(OH)2 is much higer than the mass of Cu metal. So, if we just measure the equivalent mass of metal by deposition, we can see this to increase than the normal equivalent mass expected. Because we will not able to separate the impurity from the actual.