Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
Matter is never destroyed, nor created. So, this is false.
Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.
We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y .......> equation I
We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) ............> equation II
Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml
Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml
Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml
Hope this helps :)
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