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The metal-insoluble salt electrode consi...

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:

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`

Text Solution

AI Generated Solution

The correct Answer is:
To find the maximum work done by the cell under standard conditions, we can follow these steps: ### Step 1: Identify the Cell Representation The cell representation is given as: \[ M(s) | M^{2+}(aq) || Ag^{+}(aq) | Ag(s) \] ### Step 2: Write Down the Given Standard Reduction Potentials We have the following standard reduction potentials: - For the metal M: \[ E^\circ_{M/M^{2+}} = +2.37 \, \text{V} \] - For the silver-silver chloride electrode: \[ E^\circ_{Ag/Ag^{+}} = +0.80 \, \text{V} \] ### Step 3: Determine the Half-Cell Reactions The half-cell reactions are: - Oxidation at the anode (Metal M): \[ M(s) \rightarrow M^{2+}(aq) + 2e^- \] - Reduction at the cathode (Silver): \[ Ag^{+}(aq) + e^- \rightarrow Ag(s) \] ### Step 4: Calculate the Number of Electrons Transferred (n) From the half-cell reactions, we see that: - 2 electrons are transferred in the oxidation of M. - 1 electron is involved in the reduction of Ag. To balance the electrons, we will need to multiply the silver half-reaction by 2: \[ 2Ag^{+}(aq) + 2e^- \rightarrow 2Ag(s) \] Thus, \( n = 2 \). ### Step 5: Calculate the Standard Cell Potential (E°cell) The standard cell potential is calculated as: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] Where: - \( E^\circ_{cathode} = E^\circ_{Ag/Ag^{+}} = +0.80 \, \text{V} \) - \( E^\circ_{anode} = E^\circ_{M/M^{2+}} = +2.37 \, \text{V} \) Thus, \[ E^\circ_{cell} = 0.80 \, \text{V} - 2.37 \, \text{V} = -1.57 \, \text{V} \] ### Step 6: Calculate the Maximum Work Done (ΔG) The maximum work done by the cell can be calculated using the formula: \[ \Delta G = -nFE^\circ_{cell} \] Where: - \( F = 96500 \, \text{C/mol} \) (Faraday's constant) - \( n = 2 \) - \( E^\circ_{cell} = -1.57 \, \text{V} \) Substituting the values: \[ \Delta G = -2 \times 96500 \, \text{C/mol} \times (-1.57 \, \text{V}) \] \[ \Delta G = 2 \times 96500 \times 1.57 \] \[ \Delta G = 303,610 \, \text{J} \] ### Step 7: Convert to Kilojoules To convert from joules to kilojoules: \[ \Delta G = 303.61 \, \text{kJ} \] ### Final Answer The maximum work done by the cell under standard conditions is approximately: \[ \Delta G \approx 303.61 \, \text{kJ} \] ---
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