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For the electrochemical cell, (M)|M^(+))...

For the electrochemical cell, `(M)|M^(+))||(X^(-)|X)`.
`E^(@)(M^(+)//M)=0.44V` and `E^(@)(X//X^(-))=0.33V`.
From this data one can deduce that

A

` M + X to M^ ( + ) + X^ ( -) ` is the sponthaneous reaction

B

`M^(+)+X^(-)to M+X` is the spontaneous reaction

C

`E_("cell")=0.77V`

D

`E_("cell)=-0.77V`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the electrochemical cell given by the notation `(M)|M^(+))||(X^(-)|X)`. We are provided with the standard reduction potentials for the half-reactions involving M and X. ### Step-by-Step Solution: 1. **Identify the half-reactions and their standard reduction potentials:** - For the half-reaction \( M^{+} + e^{-} \rightarrow M \), the standard reduction potential \( E^{\circ}(M^{+}/M) = 0.44 \, V \). - For the half-reaction \( X^{-} + e^{-} \rightarrow X \), the standard reduction potential \( E^{\circ}(X/X^{-}) = 0.33 \, V \). 2. **Determine the oxidation and reduction processes:** - In the electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode. - Since \( M^{+} \) has a higher reduction potential than \( X^{-} \), \( M \) will be oxidized to \( M^{+} \) (anode), and \( X^{-} \) will be reduced to \( X \) (cathode). 3. **Write the overall cell reaction:** - The overall cell reaction can be written as: \[ M + X^{-} \rightarrow M^{+} + X \] 4. **Calculate the standard cell potential \( E^{\circ}_{cell} \):** - The standard cell potential is calculated using the formula: \[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \] - Here, \( E^{\circ}_{cathode} = E^{\circ}(X/X^{-}) = 0.33 \, V \) and \( E^{\circ}_{anode} = E^{\circ}(M^{+}/M) = 0.44 \, V \). - Plugging in the values: \[ E^{\circ}_{cell} = 0.33 \, V - 0.44 \, V = -0.11 \, V \] 5. **Interpret the result:** - Since \( E^{\circ}_{cell} \) is negative, the reaction is non-spontaneous under standard conditions. ### Conclusion: From the analysis, we can deduce that the correct option is **B**, as it describes the spontaneous reaction correctly based on the calculated cell potential.

To solve the problem, we need to analyze the electrochemical cell given by the notation `(M)|M^(+))||(X^(-)|X)`. We are provided with the standard reduction potentials for the half-reactions involving M and X. ### Step-by-Step Solution: 1. **Identify the half-reactions and their standard reduction potentials:** - For the half-reaction \( M^{+} + e^{-} \rightarrow M \), the standard reduction potential \( E^{\circ}(M^{+}/M) = 0.44 \, V \). - For the half-reaction \( X^{-} + e^{-} \rightarrow X \), the standard reduction potential \( E^{\circ}(X/X^{-}) = 0.33 \, V \). ...
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