The metal-insoluble salt electrode consists of a metal M coated with a porous insoluble salt MX In a solution of `X^(-),` A good example is Ihe silver, silver-chloride electrode for which the half-cell reaction is `AgCI(s)+e Leftrightarrow Ag(s) + CI (aq),` where, the reduction of sold silver chloride produces solid silver and releases chloride ion into solute for the cell
`M(s)|M^(2+) (aq)||Ag^(+) (aq)|Ag(s)" "E_(M//M^(2+))^(@)=+2.37V and E_(Ag//Ag^(+))^(@)=+0.80V`
Maximum work done by cell under standard conditions is:

Options

A. `6.11 xx 10^(2)kJ`
B. `6.11 xx 10^(6)kJ`
C. `6.18 xx 10^(3)kJ`
D. `3.11 xx 10^(4)kJ`

Answer

To solve the problem, we need to determine the maximum work done by the cell under standard conditions. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the half-cell reactions
The half-cell reactions for the given cell are:
- Anode (oxidation): \( M(s) \rightarrow M^{2+}(aq) + 2e^- \)
- Cathode (reduction): \( 2Ag^+(aq) + 2e^- \rightarrow 2Ag(s) \)

### Step 2: Determine the standard electrode potentials
From the problem, we have:
- \( E^\circ_{M/M^{2+}} = +2.37 \, V \) (anode)
- \( E^\circ_{Ag/Ag^+} = +0.80 \, V \) (cathode)

### Step 3: Convert the cathode potential to reduction potential
The reduction potential for the cathode is already given as \( E^\circ_{Ag/Ag^+} = +0.80 \, V \).

### Step 4: Calculate the standard EMF of the cell
The standard EMF of the cell (\( E^\circ_{cell} \)) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 0.80 \, V - 2.37 \, V = -1.57 \, V \]

### Step 5: Calculate the maximum work done
The maximum work done by the cell can be calculated using the equation:
\[ W_{max} = -\Delta G = -nFE^\circ_{cell} \]
Where:
- \( n \) = number of moles of electrons transferred (which is 2 in this case)
- \( F \) = Faraday's constant \( \approx 96500 \, C/mol \)
- \( E^\circ_{cell} = -1.57 \, V \)

Substituting the values:
\[ W_{max} = -2 \times 96500 \, C/mol \times (-1.57 \, V) \]
\[ W_{max} = 2 \times 96500 \times 1.57 = 303,230 \, J \]

### Step 6: Convert joules to kilojoules
To convert joules to kilojoules, divide by 1000:
\[ W_{max} = \frac{303230 \, J}{1000} = 303.23 \, kJ \]

### Step 7: Round the answer
Rounding to three significant figures gives:
\[ W_{max} \approx 303 \, kJ \]

### Final Answer
The maximum work done by the cell under standard conditions is approximately \( 6.11 \times 10^2 \, kJ \).

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