Calculate the quantity of electricity dellvered by a Daniel celi initially containing 1 L each of 1M `Cu^(2+)` ion and 1M `Zn^(2+)` C which is operated until potential drops to 1V. Given `E^(@) Zn^(2+)//Zn=-0.76V, E_(Cu^(2+)//Cu)^(@)=+0.34V`

Answer

To solve the problem of calculating the quantity of electricity delivered by a Daniel cell, we can follow these steps:

### Step 1: Calculate the standard cell potential (E°cell)

The standard cell potential can be calculated using the formula:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]

Given:
- \( E°_{Cu^{2+}/Cu} = +0.34V \) (cathode)
- \( E°_{Zn^{2+}/Zn} = -0.76V \) (anode)

Substituting the values:
\[ E°_{cell} = 0.34V - (-0.76V) \]
\[ E°_{cell} = 0.34V + 0.76V \]
\[ E°_{cell} = 1.10V \]

### Step 2: Write the half-reactions

At the anode (oxidation):
\[ Zn \rightarrow Zn^{2+} + 2e^- \]

At the cathode (reduction):
\[ Cu^{2+} + 2e^- \rightarrow Cu \]

### Step 3: Write the overall cell reaction

Combining the half-reactions gives us:
\[ Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu \]

### Step 4: Use the Nernst equation to find the conditions when Ecell = 1V

The Nernst equation is given by:
\[ E_{cell} = E°_{cell} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) \]

Where:
- \( n = 2 \) (number of electrons transferred)

Setting \( E_{cell} = 1V \):
\[ 1V = 1.10V - \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) \]

Rearranging gives:
\[ \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) = 1.10V - 1V \]
\[ \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) = 0.10V \]

Now, multiplying both sides by \( \frac{2}{0.059} \):
\[ \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) = \frac{0.10 \times 2}{0.059} \]
\[ \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) \approx 3.3898 \]

### Step 5: Solve for concentrations

This implies:
\[ \frac{[Zn^{2+}]}{[Cu^{2+}]} = 10^{3.3898} \]
Calculating this gives:
\[ [Zn^{2+}] \approx 2451.2 \times [Cu^{2+}] \]

### Step 6: Determine the change in concentration

Let \( x \) be the amount of \( Zn \) oxidized and \( Cu^{2+} \) reduced:
- Initial concentrations: \( [Zn^{2+}] = 0 \) and \( [Cu^{2+}] = 1 \)
- After reaction: \( [Zn^{2+}] = x \) and \( [Cu^{2+}] = 1 - x \)

From the ratio:
\[ x = 2451.2(1 - x) \]
Solving this gives:
\[ x + 2451.2x = 2451.2 \]
\[ 2452.2x = 2451.2 \]
\[ x \approx 0.9996 \]

### Step 7: Calculate the quantity of electricity

The quantity of electricity (Q) can be calculated using:
\[ Q = n \times F \times moles \]

Where:
- \( n = 2 \) (from the balanced equation)
- \( F = 96500 \, C/mol \)
- Moles of \( Zn \) oxidized = \( 0.9996 \, mol \)

Thus:
\[ Q = 2 \times 96500 \times 0.9996 \]
\[ Q \approx 192249.2 \, C \]

### Final Answer
The quantity of electricity delivered by the Daniel cell is approximately **192249.2 Coulombs**.

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