In the electrochemistry lab we assumed that the electrode process at the copper
ID: 1020266 • 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
answers :
Cu(OH)2 solid is formed when water reduced at cathod
equivalent mass for the unknown metal remain the same
explanation:
actual reactions
anode reaction : H2(g) + 2OH- (aq) -------------------> 2H20 + 2e-
cathode reaction : Cu+2 + 2e- --------------------> Cu (s)
but at cathode some of Cu+2 react with OH- forms Cu(OH)2 solid
Cu+2 + 2OH- ---------------->Cu(OH)2 (s)
but Cu equivalent mass doesnot change it is constant . becuase molar mass of Cu is fixed . the charge is also fixed +2
so equivalent mass = molar mass / 2 = 63.5 /2 = 31.75 . it is fixed