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Standard reduction potentials at 298 K: Pb2+/Pb = -0.13 V, Ni2+/Ni = -0.24 V, Cd2+/Cd = -0.40 V, Fe2+/Fe = -0.44 V. A solution contains 0.001 M of metal X2+ and 0.1 M of metal Y2+. Metal rods X and Y are inserted and connected by a wire. Metal X dissolves. Which combination of X and Y (respectively) is/are correct? (R = 8.314 J/K/mol, F = 96500 C/mol)

1. Cd and Ni
2. Cd and Fe
3. Ni and Pb
4. Ni and Fe

Correct answer: Cd and Ni

Solution

X dissolves (anode). E_X < E_Y required. Apply Nernst: E = E_std + (0.0296/2)*log[M2+] (for 2e- process at 298 K). For Cd at 0.001 M: E_Cd = -0.40 + 0.0148*log(0.001) = -0.40 - 0.0444 = -0.4444 V. For Ni at 0.1 M: E_Ni = -0.24 + 0.0148*(-1) = -0.2548 V. E_cell = -0.2548 - (-0.4444) = +0.19 V > 0. Cd dissolves. Valid. For Ni and Pb: E_Ni(0.001M) = -0.2844 V, E_Pb(0.1M) = -0.1448 V. E_cell = +0.14 V > 0. Ni dissolves. Valid. Options (A) and (C) are correct.

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