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

Answer: (C)

M$^{2+}$ + X$^{2-}$ $\rightarrow$ M + X is a spontaneous reaction

Answer 2

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).