A fuel cell is a cell that is continously supplied with an oxidant and a reductant so that if can deliver a current indefinitely.
Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work.Internal combustion engines at best convert only the fraction `(T_2-T_1)//T_2` of the heat of combustion into mechanical work.
While the thermodynamic efficiency of the fuel cell is given by, `eta=(DeltaG)/(DeltaH)`, where `DeltaG` is the Gibbs energy change for the cell reaction and `DeltaH` is the enthalpy change of the cell reaction.A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte.
`Pt|H_2(g)|H^(+)(aq.)||H_2O(l)|O_(2)(g)|Pt , (2.303 RT)/F=0.06`
The above fuel cell is used to produce constant current supply under constant temperature & 30 atm constant total pressure conditions in a cylinder.If 10 moles `H_2` and 5 moles of `O_2` were taken initially.
Rate of consumption of `O_2` is 10 milli moles per minute. The half-cell reactions are
`1/2O_2(g)+2H^+(aq)+2e^(-)toH_2O(l) E^(@)=1.246 V`
`2H^+(aq)+2e^(-) to H_2(g) E^(@)=0`
To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized.Since fused salts have lower electolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications.
Calculate e.m.f of the given cell at t=0.(log 2=0.3)

Options

A. `1.255 V`
B. `1.35 V`
C. `1.3 V`
D. `1.246 V`

Explanation

To calculate the e.m.f (electromotive force) of the given hydrogen-oxygen fuel cell at time \( t = 0 \), we will use the Nernst equation. Let's go through the steps one by one.

### Step 1: Write the Nernst Equation
The Nernst equation is given by:

\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln \left( \frac{[\text{Oxidized}]}{[\text{Reduced}]} \right) \]

Where:
- \( E_{\text{cell}} \) is the cell potential.
- \( E^{\circ}_{\text{cell}} \) is the standard cell potential.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons transferred in the reaction.
- \( F \) is Faraday's constant (96485 C/mol).
- The ratio of concentrations or pressures of the oxidized and reduced forms is used.

### Step 2: Determine the Standard Cell Potential
From the half-cell reactions provided:
1. \( \frac{1}{2} O_2(g) + 2H^+(aq) + 2e^- \rightarrow H_2O(l) \quad E^{\circ} = 1.246 \, V \)
2. \( 2H^+(aq) + 2e^- \rightarrow H_2(g) \quad E^{\circ} = 0 \, V \)

The standard cell potential \( E^{\circ}_{\text{cell}} \) can be calculated as:

\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.246 \, V - 0 \, V = 1.246 \, V \]

### Step 3: Calculate the Number of Electrons Transferred
From the half-cell reactions, we see that 2 moles of electrons are transferred in the overall reaction:

\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \]

Thus, \( n = 2 \).

### Step 4: Calculate the Pressures of Reactants
We know the total pressure is 30 atm, and we have 10 moles of \( H_2 \) and 5 moles of \( O_2 \).

Using the ideal gas law, we can find the partial pressures:
- Total moles = \( 10 + 5 = 15 \) moles
- Pressure of \( H_2 \) = \( \frac{10}{15} \times 30 \, \text{atm} = 20 \, \text{atm} \)
- Pressure of \( O_2 \) = \( \frac{5}{15} \times 30 \, \text{atm} = 10 \, \text{atm} \)

### Step 5: Substitute Values into the Nernst Equation
Now we can substitute the values into the Nernst equation:

\[ E_{\text{cell}} = 1.246 \, V - \frac{(0.06)}{2} \log \left( \frac{1}{20 \times \sqrt{10}} \right) \]

### Step 6: Calculate the Logarithmic Term
Calculating the logarithmic term:

\[ \frac{1}{20 \times \sqrt{10}} = \frac{1}{20 \times 3.162} \approx \frac{1}{63.24} \approx 0.0158 \]

Now, calculate the logarithm:

\[ \log(0.0158) \approx -1.8 \quad (\text{since } \log(2) = 0.3 \text{ and } \log(10) = 1) \]

### Step 7: Final Calculation
Substituting back into the Nernst equation:

\[ E_{\text{cell}} = 1.246 \, V - \frac{(0.06)}{2} \times (-1.8) \]

Calculating:

\[ E_{\text{cell}} = 1.246 \, V + 0.054 = 1.300 \, V \]

### Final Answer
Thus, the e.m.f of the given cell at \( t = 0 \) is approximately:

\[ \boxed{1.3 \, V} \]