The voltage of a certain cell has standard potential at `25^(@)C` and `20^(@)C` are 0.3525 V and 0.3533 V respectively. If the number of electrons involved in te overall reactions are two, calculate `triangleG^(@), triangleS^(@)` and `triangleH^(@)` at `25^(@)C`.

Explanation

To solve the problem step by step, we will calculate the standard Gibbs free energy change (ΔG°), standard entropy change (ΔS°), and standard enthalpy change (ΔH°) at 25°C using the given standard potentials at two different temperatures.

### Step 1: Calculate the change in standard potential (dE/dT)

We know the standard potentials at two temperatures:
- E° at 25°C = 0.3525 V
- E° at 20°C = 0.3533 V

Using the formula for the change in standard potential with temperature:
\[ \frac{dE}{dT} = \frac{E°_{25°C} - E°_{20°C}}{T_2 - T_1} \]
Substituting the values:
\[ \frac{dE}{dT} = \frac{0.3525 - 0.3533}{25 - 20} = \frac{-0.0008}{5} = -0.00016 \, \text{V/°C} \]

### Step 2: Calculate ΔG° at 25°C

The relationship between Gibbs free energy change and standard potential is given by:
\[ \Delta G° = -nFE° \]
Where:
- n = number of electrons = 2
- F = Faraday's constant = 96500 C/mol
- E° = E° at 25°C = 0.3525 V

Substituting the values:
\[ \Delta G° = -2 \times 96500 \times 0.3525 = -68.03 \, \text{kJ/mol} \]

### Step 3: Calculate ΔS° at 25°C

Using the relationship derived from the Gibbs free energy equation:
\[ \Delta G° = \Delta H° - T\Delta S° \]
We can rearrange it to find ΔS°:
\[ \Delta S° = \frac{\Delta H° - \Delta G°}{T} \]
However, we need ΔH° to calculate ΔS°.

Instead, we can calculate ΔS° directly using the formula:
\[ \Delta S° = -\frac{dE}{dT} \cdot \frac{nF}{T} \]
Substituting the values:
\[ \Delta S° = -(-0.00016) \cdot \frac{2 \times 96500}{298} \]
Calculating:
\[ \Delta S° = 0.00016 \cdot \frac{193000}{298} \approx 0.1 \, \text{kJ/K} \approx 100 \, \text{J/K} \]

### Step 4: Calculate ΔH° at 25°C

Now we can use the Gibbs-Helmholtz equation:
\[ \Delta H° = \Delta G° + T\Delta S° \]
Substituting the values:
\[ \Delta H° = -68.03 + 298 \cdot 0.1 \]
Calculating:
\[ \Delta H° = -68.03 + 29.8 = -38.23 \, \text{kJ/mol} \]

### Final Results

- ΔG° = -68.03 kJ/mol
- ΔS° = 100 J/K (or 0.1 kJ/K)
- ΔH° = -38.23 kJ/mol