Question

For the electrochemical cell
M|M$^{2+}$||X$^{2-}$|X
If $E^0_{(M^{2+}/M)} = 0.46$ V and $E^0_{(X/X^{2-})} = 0.34$ V.
Which of the following is correct?

• A. $E_{cell} = -0.80$ V
• B. M + X $\rightarrow$ M$^{2+}$ + X$^{2-}$ is a spontaneous reaction
• C. M$^{2+}$ + X$^{2-}$ $\rightarrow$ M + X is a spontaneous reaction
• D. $E_{cell} = 0.80$ V}

Answer 1

The Correct Option is C

The standard cell potential \(E^\circ_{\text{cell}}\) is calculated as:
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}.\]
Step 1: Identify the anode and cathode
- \(E^\circ (\text{M}^{2+}/\text{M}) = 0.46 \, \text{V}\),
- \(E^\circ (\text{X}/\text{X}^{2-}) = 0.34 \, \text{V}\).
Since \(\text{M}^{2+}/\text{M}\) has a higher reduction potential, it will act as the cathode, and \(\text{X}/\text{X}^{2-}\) will act as the anode.
Step 2: Calculate \(E^\circ_{\text{cell}}\) 
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - 0.46 = -0.12 \, \text{V}.\]
Step 3: Analyze the spontaneity of the reaction 
Since \(E^\circ_{\text{cell}}\) is negative, the reaction will proceed in the reverse direction (the reverse reaction is spontaneous).  
The spontaneous reaction is:
\[\text{M}^{2+} + \text{X}^{2-} \rightarrow \text{M} + \text{X}.\]
Step 4: Validate the options
-Option (1): Incorrect, as \(E_{\text{cell}} = -0.12 \, \text{V}\), not \(-0.80 \, \text{V}\).  
-Option (2): Incorrect, as \(\text{M} + \text{X}^{2-} \rightarrow \text{M}^{2+} + \text{X}^{2-}\) is not spontaneous.  
-Option (3): Correct, as \(\text{M}^{2+} + \text{X}^{2-} \rightarrow \text{M} + \text{X}\) is the spontaneous reaction.  
-Option (4): Incorrect, as \(E_{\text{cell}} = -0.12 \, \text{V}\), not \(0.80 \, \text{V}\).
Final Answer: (3).

Answer 2

Step 1: Given data.
Standard reduction potentials are:
E°(M²⁺/M) = +0.46 V
E°(X/X²⁻) = +0.34 V

Step 2: Identify the cathode and anode.
The species with a higher reduction potential acts as the cathode (gets reduced), while the one with a lower reduction potential acts as the anode (gets oxidized).

Since 0.46 V > 0.34 V,
- M²⁺/M acts as the cathode (reduction occurs here).
- X/X²⁻ acts as the anode (oxidation occurs here).

Step 3: Write the half-cell reactions.
At cathode (reduction): M²⁺ + 2e⁻ → M
At anode (oxidation): X²⁻ → X + 2e⁻

Step 4: Overall cell reaction.
By combining the two half-reactions:
M²⁺ + X²⁻ → M + X

Step 5: Check spontaneity.
Cell potential, E°_cell = E°_cathode - E°_anode
E°_cell = 0.46 - 0.34 = +0.12 V
Since E°_cell is positive, the reaction is spontaneous.

Step 6: Final Answer.
The spontaneous reaction is:
M²⁺ + X²⁻ → M + X

Final Answer: M²⁺ + X²⁻ → M + X is a spontaneous reaction.