Manohar Sehgal
By adding a few drops of 0.1M HCl , the following reaction takes place:
Cu(s) + H2O2 (aq) + 2 H (+I) = Cu (+2) + 2 H2O(l) -----[1]
As It has a positive value of cell voltage [1.44 V] with the following half cell potentials
Cu(s)=Cu(+2) +2e E0= -0.34 V
H2O2 (aq) + 2H (+I) +2e=2H2O E0=+1.78 V
On the other hand Cu (I) were to form then
Cu(s) =Cu (I) +e ------------ E0= - 0.522 V
The cell voltage of the reaction
2Cu(s) + H2O2 (aq) + 2 H( +I) = 2Cu (I) + 2 H2O(l) ------[2]
Should be +1.78-0.522=1.258 V; i.e less than the first reaction & hence the reacton [1]
is all the more feasible reaction .So Cu (II) compound should form.
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