The standard cell potential of:
`Zn(s) abs( Zn^(2+) (aq) )abs( Cu^(2+) (aq)) Cu (s)` cell is `1.10` V The maximum work obtained by this cell will be

Options

A. `–106.15` kJ
B. `– 212.30` kJ
C. `–318.45` kJ
D. `– 424.60` kJ

Answer

To find the maximum work obtained by the given electrochemical cell, we can use the formula:

\[ W_{\text{max}} = -nFE^\circ_{\text{cell}} \]

where:
- \( n \) = number of moles of electrons transferred in the reaction,
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \)),
- \( E^\circ_{\text{cell}} \) = standard cell potential (given as \( 1.10 \, \text{V} \)).

### Step 1: Identify the number of electrons transferred (n)
From the given cell reaction:
\[ \text{Zn} (s) + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu} (s) \]
We see that zinc is oxidized to zinc ions, losing 2 electrons, and copper ions are reduced to copper, gaining 2 electrons. Therefore, the number of electrons transferred (\( n \)) is 2.

### Step 2: Use Faraday's constant (F)
The value of Faraday's constant is:
\[ F = 96500 \, \text{C/mol} \]

### Step 3: Substitute the values into the formula
Now substituting the values into the formula for maximum work:
\[ W_{\text{max}} = -nFE^\circ_{\text{cell}} = - (2) (96500 \, \text{C/mol}) (1.10 \, \text{V}) \]

### Step 4: Calculate the work done
Calculating the above expression:
\[ W_{\text{max}} = - (2) (96500) (1.10) = -212300 \, \text{J} \]

### Step 5: Convert joules to kilojoules
To convert joules to kilojoules, we divide by 1000:
\[ W_{\text{max}} = -212300 \, \text{J} \div 1000 = -212.3 \, \text{kJ} \]

### Final Answer
The maximum work obtained by this cell is:
\[ W_{\text{max}} = -212.3 \, \text{kJ} \]

---