Calculate the e.m.f. of the cell
`Zn//Zn^(2+) (0.1M) "||" Cu^(2+) (0.01M) "|"Cu`
`E_(Zn^(2+)//Zn)^(Theta)= -0.76 " V and "E_(Cu^(2+)//Cu)^(Theta)=0.34V`

Explanation

To calculate the e.m.f. (electromotive force) of the cell represented as `Zn//Zn^(2+) (0.1M) "||" Cu^(2+) (0.01M) "|"Cu`, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials
- The half-reaction for zinc (anode) is:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \quad (E^\circ = -0.76 \, \text{V}) \]
- The half-reaction for copper (cathode) is:
\[ \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \quad (E^\circ = 0.34 \, \text{V}) \]

### Step 2: Calculate the standard cell potential (E°cell)
The standard cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V} \]

### Step 3: Use the Nernst equation to calculate the cell potential (Ecell)
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Here, \(n\) is the number of moles of electrons transferred, which is 2 for this reaction.

### Step 4: Determine the concentrations of products and reactants
In our case:
- Products: \([\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- Reactants: \([\text{Cu}^{2+}] = 0.01 \, \text{M}\)

### Step 5: Substitute into the Nernst equation
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 1.10 \, \text{V} - \frac{0.0591}{2} \log \left( \frac{0.1}{0.01} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{0.1}{0.01} \right) = \log(10) = 1 \]
Now substituting this back into the equation:
\[ E_{\text{cell}} = 1.10 \, \text{V} - \frac{0.0591}{2} \cdot 1 \]
\[ E_{\text{cell}} = 1.10 \, \text{V} - 0.02955 \, \text{V} = 1.07045 \, \text{V} \]

### Step 6: Round the final answer
Rounding to three significant figures:
\[ E_{\text{cell}} \approx 1.07 \, \text{V} \]

### Final Answer
The e.m.f. of the cell is approximately **1.07 V**.

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