Identify the given standard reduction potentials

The cell is written as $\mathrm{M}|\mathrm{M}^{2+}||\mathrm{X}|\mathrm{X}^{2-}$

Given: $E^\circ_{(\mathrm{M}^{2+}/\mathrm{M})}=0.46\,\text{V}$ $E^\circ_{(\mathrm{X}/\mathrm{X}^{2-})}=0.34\,\text{V}$

Here, both are written as reduction potentials: $\mathrm{M}^{2+}+2e^- \rightarrow \mathrm{M} \qquad E^\circ=0.46\,\text{V}$ $\mathrm{X}+2e^- \rightarrow \mathrm{X}^{2-} \qquad E^\circ=0.34\,\text{V}$

Decide cathode and anode

The half-cell with the higher reduction potential undergoes reduction at the cathode.

Since $0.46\,\text{V} > 0.34\,\text{V}$ so:

• Cathode: $\mathrm{M}^{2+}+2e^- \rightarrow \mathrm{M}$
• Anode: reverse of the X half-reaction: $\mathrm{X}^{2-} \rightarrow \mathrm{X}+2e^-$

Write the overall cell reaction

Adding the two half-reactions: $\mathrm{M}^{2+}+2e^- \rightarrow \mathrm{M}$ $\mathrm{X}^{2-} \rightarrow \mathrm{X}+2e^-$

Overall: $\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}$

Calculate standard cell potential

$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$

Using reduction potentials: $E^\circ_{\text{cell}}=0.46-0.34=0.12\,\text{V}$

Since this is positive, the reaction $\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}$ is spontaneous.

Check each option

• A: $E_{\text{cell}}=0.80\,\text{V}$

  Incorrect, because $E^\circ_{\text{cell}}=0.12\,\text{V}$.

• B: $\mathrm{M}+\mathrm{X} \rightarrow \mathrm{M}^{2+}+\mathrm{X}^{2-}$ is spontaneous

  Incorrect. This is the reverse reaction, so it is non-spontaneous under standard conditions.

• C: $E_{\text{cell}}=-0.80\,\text{V}$

  Incorrect. The value is neither negative nor equal to $0.80\,\text{V}$.

• D: $\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}$ is a spontaneous reaction

  Correct.

Final answer

The correct option is: $\boxed{\text{D}}$

Note: The spontaneous reaction is correct, but the numerical options A and C appear inconsistent with the given potentials. The correct cell emf from the given data is $0.12\,\text{V}$, not $0.80\,\text{V}$.