When a rod of metal A is dipped in an aqueous solution of metal B (concentration of B²⁺ ion being 1M) at 25°C, the standard electrode potentials are A²⁺ | A = −0.761 and B²⁺ | B = +0.34 V.

B will gradually deposit on A

The correct answer is option 2. B will gradually deposit on A.

To determine the outcome of the reaction between metal A and metal B in an aqueous solution, we can compare their standard electrode potentials. Here, metal A is more reactive than metal B because the standard electrode potential of A is more negative than that of B.

Given:

Standard electrode potential of A: \( E^\circ_{\text{A}^{2+}/\text{A}} = -0.761 \) V

Standard electrode potential of B: \( E^\circ_{\text{B}^{2+}/\text{B}} = +0.34 \) V

Since metal A has a more negative standard electrode potential than metal B, it is more likely to undergo oxidation compared to metal B. Therefore, metal A will act as the anode and undergo oxidation, while metal B will act as the cathode and undergo reduction.

Reaction at the Anode (Metal A): \(\text{A} \rightarrow \text{A}^{2+} + 2e^- \)

Reaction at the Cathode (Metal B): \(\text{B}^{2+} + 2e^- \rightarrow \text{B} \)

Overall Cell Reaction: \(\text{A} + \text{B}^{2+} \rightarrow \text{A}^{2+} + \text{B}\)

Since metal A undergoes oxidation and metal B undergoes reduction, metal A will gradually dissolve into the aqueous solution as \( \text{A}^{2+} \) ions, while metal B will gradually deposit onto metal A.

So, the correct option is: B will gradually deposit on A.