Question

The electronic configuration of Cu(II) is 3d9 whereas that of Cu(I) is 3d10. Which of the following is correct in an aqueous medium?

a. Cu(II) is reducing
b. Cu(I) is reducing
c. Cu(II) is oxidizing
d. Both Cu(I) and Cu(II) are stable

Answer 1

In an aqueous medium, Cu(II) is oxidizing while Cu(I) is reducing.

Step-by-step explanation:

In an aqueous medium, Cu(II) is oxidizing while Cu(I) is reducing.

In the Cu(II) oxidation state, the copper ion gains two electrons to form Cu(II). This can be seen in the equation Cu(s) → Cu²+ (aq) + 2e¯, where Cu²+ is formed.

On the other hand, in the Cu(I) reduction state, the copper ion loses one electron to form Cu(I). This can be seen in the equation 2 × (Ag+ (aq) + e¯ → Ag(s)), where Cu(I) is formed.

Therefore, option b. Cu(I) is reducing is correct.