For the electrochemical cell, if E°(M2+/M) = 0.46\ V and E°(X/X2-) = 0.34\ V. Which of the following is correct?

Strategy: For an electrochemical cell to be spontaneous, the standard cell potential ($E^\circ_{\text{cell}}$) must be positive. This is achieved when the half-cell with the higher reduction potential acts as the cathode.

Step 1: Compare reduction potentials

• $E^\circ_{\ce{M^{2+}/M}} = +0.46\text{ V}$
• $E^\circ_{\ce{X/X^{2-}}} = +0.34\text{ V}$

Step 2: Determine cathode and anode for spontaneity Cathode (Higher $E^\circ$) = $\ce{M^{2+}/M}$ Anode (Lower $E^\circ$) = $\ce{X/X^{2-}}$ $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.46 - 0.34 = +0.12\text{ V}$

Step 3: Write the spontaneous reaction Reduction at Cathode: $\ce{M^{2+} + 2e^- \rightarrow M}$ Oxidation at Anode: $\ce{X^{2-} \rightarrow X + 2e^-}$ Net Spontaneous Reaction: $\ce{M^{2+} + X^{2-} \rightarrow M + X}$

$\boxed{\text{Answer: (d) } \ce{M^{2+} + X^{2-} \rightarrow M + X} \text{ is a spontaneous reaction}}$