The standard reduction potential of `Zn^(2+)|Zn` and `Cu^(2+)|Cu` are `-0.76V` and `+0.34` V respectively. What is the cell e.m.f (in V) of the following cell?
`((RT)/(F)=0.059)`
`Zn//Zn^(2+)" "(0.05M)//Cu^(2+)(0.005M)//Cu`

Options

A. `1.1295`
B. `1.0705`
C. `1.1`
D. `1.041`

Explanation

To calculate the cell e.m.f (electromotive force) for the given electrochemical cell `Zn//Zn^(2+)(0.05M)//Cu^(2+)(0.005M)//Cu`, we will follow these steps:

### Step 1: Identify the half-reactions
The standard reduction potentials are given as:
- For `Zn^(2+) + 2e^- → Zn`, \( E^\circ = -0.76 \, V \)
- For `Cu^(2+) + 2e^- → Cu`, \( E^\circ = +0.34 \, V \)

In this cell:
- Zinc is oxidized (anode): `Zn → Zn^(2+) + 2e^-`
- Copper is reduced (cathode): `Cu^(2+) + 2e^- + 2e^- → Cu`

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

### Step 3: Calculate the reaction quotient (Q)
The reaction quotient \( Q \) for the cell reaction is given by:
\[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \]
Substituting the concentrations:
\[ Q = \frac{0.05}{0.005} = 10 \]

### Step 4: Apply the Nernst equation
The Nernst equation is:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \]
Given that \( \frac{RT}{F} = 0.059 \, V \) and \( n = 2 \) (since 2 electrons are involved), we can rewrite the equation as:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log Q \]

### Step 5: Substitute values into the Nernst equation
Substituting the known values:
\[ E_{cell} = 1.10 \, V - \frac{0.059}{2} \log(10) \]
Since \( \log(10) = 1 \):
\[ E_{cell} = 1.10 \, V - \frac{0.059}{2} \]
\[ E_{cell} = 1.10 \, V - 0.0295 \, V \]
\[ E_{cell} = 1.0705 \, V \]

### Step 6: Final answer
The cell e.m.f is:
\[ E_{cell} \approx 1.07 \, V \]