Equation: 3Cu(s) 8HNO3(aq) --> 2NO(g) 3Cu(NO3)2(aq) 4H2O(l) In the above reaction, the element oxidized is , the reducing agent is and the number of electrons transferred from reducing to oxidizing agent in the equation, as written, is .

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.

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.