Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as:
`6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)`
The energy of one mole of a photon of wave lenght is known as one Einstein.
Calculate the emf of this fuel cell at `25^(@)` C if `DeltaH^(@)` and `DeltaS^(@)` of photosynthesis are `2.82xx10^(6) "J mol"^(-1)` and -182 `J"mol"^(-1)` respectively.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the standard Gibbs free energy change (ΔG°)
The formula for standard Gibbs free energy change is given by:

\[ \Delta G° = \Delta H° - T \Delta S° \]

Where:
- ΔH° = 2.82 × 10^6 J/mol (enthalpy change)
- ΔS° = -182 J/mol·K (entropy change)
- T = 25°C = 25 + 273 = 298 K (temperature in Kelvin)

Substituting the values into the equation:

\[ \Delta G° = 2.82 \times 10^6 \, \text{J/mol} - (298 \, \text{K} \times -182 \, \text{J/mol·K}) \]

Calculating the second term:

\[ 298 \times -182 = -54236 \, \text{J/mol} \]

Now substituting this back into the Gibbs free energy equation:

\[ \Delta G° = 2.82 \times 10^6 \, \text{J/mol} + 54236 \, \text{J/mol} \]
\[ \Delta G° = 2.874236 \times 10^6 \, \text{J/mol} \]

### Step 2: Determine the number of electrons transferred (n)
In the photosynthesis reaction, carbon goes from CO₂ to glucose (C₆H₁₂O₆).

- In CO₂, the oxidation state of carbon is +4.
- In glucose, the oxidation state of carbon is 0.

The change in oxidation state for one carbon atom is:

\[ 0 - (+4) = -4 \]

Since there are 6 carbon atoms in glucose:

\[ \text{Total change} = 6 \times 4 = 24 \text{ electrons} \]

### Step 3: Use the Gibbs free energy to find the emf (E°cell)
The relationship between Gibbs free energy and emf is given by:

\[ \Delta G° = -nFE°_{cell} \]

Where:
- F = 96500 C/mol (Faraday's constant)
- n = 24 (number of electrons)

Rearranging the equation to solve for E°cell:

\[ E°_{cell} = -\frac{\Delta G°}{nF} \]

Substituting the known values:

\[ E°_{cell} = -\frac{2.874236 \times 10^6 \, \text{J/mol}}{24 \times 96500 \, \text{C/mol}} \]

Calculating the denominator:

\[ 24 \times 96500 = 2316000 \, \text{C/mol} \]

Now substituting back into the equation:

\[ E°_{cell} = -\frac{2.874236 \times 10^6}{2316000} \]

Calculating the emf:

\[ E°_{cell} \approx -1.24 \, \text{V} \]

### Final Answer:
The emf of the fuel cell at 25°C is approximately **-1.24 V**.

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