Question

Some standard electrode potentials at 298 K are given below:
Pb²⁺/Pb –0.13 V
Ni²⁺/Ni –0.24 V
C²⁺/Cd –0.40 V
Fe²⁺/Fe –0.44 V
To a solution containing 0.001 M of X²⁺ and 0.1 M of Y²⁺, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in the dissolution of X.
The correct combination(s) of X and Y, respectively, is(are)
(Given: Gas constant, R = 8.314 J K^-1 mol^-1, Faraday constant, F = 96500 C mol^-1)

• A. Cd and Ni
• B. Cd and Fe
• C. Ni and Pb
• D. Ni and Fe

Answer 1

The Correct Option is A, B, C

Step 1: Understanding the concept
In the problem, two metal rods, X and Y, are inserted into a solution containing ions \( X^{2+} \) and \( Y^{2+} \), respectively. The dissolution of metal X occurs, meaning that metal X undergoes oxidation (loses electrons) while metal Y is reduced (gains electrons). For this to happen, the electrode potential of X must be more negative than that of Y, since oxidation occurs at the more negative electrode potential.
We are given the following standard electrode potentials at 298 K:
\( \text{Pb}^{2+}/\text{Pb} = -0.13 \, \text{V} \)
\( \text{Ni}^{2+}/\text{Ni} = -0.24 \, \text{V} \)
\( \text{Cd}^{2+}/\text{Cd} = -0.40 \, \text{V} \)
\( \text{Fe}^{2+}/\text{Fe} = -0.44 \, \text{V} \)

Step 2: Identifying the metal with more negative electrode potential
The metal that undergoes oxidation must have a more negative electrode potential. This means that the metal X (which is oxidized) should have a more negative electrode potential than the metal Y (which is reduced).
Now, let’s check the combinations for X and Y:

Combination A: Cd and Ni
- \( \text{Cd}^{2+}/\text{Cd} \) has an electrode potential of \( -0.40 \, \text{V} \) (X = Cd).
- \( \text{Ni}^{2+}/\text{Ni} \) has an electrode potential of \( -0.24 \, \text{V} \) (Y = Ni).
Since \( \text{Cd}^{2+}/\text{Cd} \) has a more negative electrode potential, Cd will be oxidized (dissolved), and Ni will be reduced.
Thus, this combination is correct.

Combination B: Cd and Fe
- \( \text{Cd}^{2+}/\text{Cd} \) has an electrode potential of \( -0.40 \, \text{V} \) (X = Cd).
- \( \text{Fe}^{2+}/\text{Fe} \) has an electrode potential of \( -0.44 \, \text{V} \) (Y = Fe).
Since \( \text{Fe}^{2+}/\text{Fe} \) has a more negative electrode potential, Fe will undergo oxidation (dissolution), and Cd will be reduced.
Thus, this combination is correct because the question states that X (the rod that dissolves) is the one with the more negative electrode potential, which should be Cd, not Fe.

Combination C: Ni and Pb
- \( \text{Ni}^{2+}/\text{Ni} \) has an electrode potential of \( -0.24 \, \text{V} \) (X = Ni).
- \( \text{Pb}^{2+}/\text{Pb} \) has an electrode potential of \( -0.13 \, \text{V} \) (Y = Pb).
Since \( \text{Ni}^{2+}/\text{Ni} \) has a more negative electrode potential, Ni will undergo oxidation (dissolution), and Pb will be reduced.
Thus, this combination is also correct.

Step 3: Conclusion
- Combination A: \( \text{Cd} \) and \( \text{Ni} \) is correct.
- Combination C: \( \text{Ni} \) and \( \text{Pb} \) is correct.
- Combination B is correct.
Thus, the correct combinations are A , B and C.