The standard electrode potentials ($$E^\circ_{M^{+}/M}$$) of four metals A, B, C and D are $$-1.2$$ V, 0.6 V, 0.85 V and $$-0.76$$ V, respectively. The sequence of deposition of metals on applying potential is:

The standard electrode potential ($$E^\circ_{M^{+}/M}$$) indicates the tendency of a metal ion to be reduced to its metal form. A higher (more positive) reduction potential means the metal ion is more likely to gain electrons and deposit as metal during electrolysis. Therefore, the metal with the highest reduction potential deposits first, followed by the next highest, and so on.

Given the standard electrode potentials:

• Metal A: $$E^\circ = -1.2 \text{V}$$
• Metal B: $$E^\circ = 0.6 \text{V}$$
• Metal C: $$E^\circ = 0.85 \text{V}$$
• Metal D: $$E^\circ = -0.76 \text{V}$$

To determine the deposition sequence, we arrange these potentials in decreasing order (from highest to lowest):

• Metal C has the highest potential: $$0.85 \text{V}$$
• Metal B has the next highest: $$0.6 \text{V}$$
• Metal D has the next: $$-0.76 \text{V}$$
• Metal A has the lowest: $$-1.2 \text{V}$$

Thus, the order of deposition is C first, followed by B, then D, and finally A. This sequence is written as C > B > D > A.

Comparing with the options:

• Option A: B > D > C > A
• Option B: A > C > B > D
• Option C: C > B > D > A
• Option D: D > A > B > C

Option C matches the sequence C > B > D > A.

Hence, the correct answer is Option C.