One ecologically important equilibrium is that between carbonate and hydrogen carbonate ions in natural water. This standard Gibb's energies of formation of `CO_(3)^(-2)(aq)` and `HCO_(3)^(Theta)(aq)` are `-527.8 kJ mol^(-1)` and `-586.8 kJ mol^(-1)` respectively.
For water,
`2H_(2)O(l) +2e^(Theta) rarr H_(2)(g) +2OH^(Theta)(aq) , E_(RP)^(Theta) =- 0.83 V`
`2H_(2)O rarr O_(2) +4H^(+) +4e^(Theta), E_(OX)^(Theta) = - 1.23V`
Calculate the potential of a cell of `pH = 7` in which the cell reaction is-
`Na_(2)CO_(3)(aq) +H_(2)O(l) rarr NaHCO_(3)(aq).+NaOH(aq)`

Options

A. `-0.22` volt
B. `+0.193` volt
C. `-0.0135` volt
D. `-0.41` volt

Answer

To solve the problem, we need to calculate the potential of a cell at pH = 7 for the reaction:

\[ \text{Na}_2\text{CO}_3(aq) + \text{H}_2\text{O}(l) \rightarrow \text{NaHCO}_3(aq) + \text{NaOH}(aq) \]

### Step 1: Identify the half-reactions
The overall reaction can be broken down into two half-reactions:
1. Reduction half-reaction:
\[ \text{CO}_3^{2-} + 2\text{H}^+ + 2e^- \rightarrow \text{HCO}_3^{-} \]
2. Oxidation half-reaction:
\[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \]

### Step 2: Calculate the Gibbs free energy change for the reaction
The Gibbs free energy change (\( \Delta G \)) for the reaction can be calculated using the standard Gibbs energies of formation:
- For \( \text{CO}_3^{2-} \): \( -527.8 \, \text{kJ/mol} \)
- For \( \text{HCO}_3^{-} \): \( -586.8 \, \text{kJ/mol} \)

Using the formula:
\[ \Delta G = \Delta G_f(\text{products}) - \Delta G_f(\text{reactants}) \]

For our reaction:
\[ \Delta G = [-586.8] - [-527.8] = -59.0 \, \text{kJ/mol} \]

### Step 3: Convert Gibbs free energy to volts
Using the relationship between Gibbs free energy and cell potential:
\[ E = -\frac{\Delta G}{nF} \]
Where:
- \( n = 2 \) (number of moles of electrons transferred)
- \( F = 96500 \, \text{C/mol} \)

Substituting the values:
\[ E = -\frac{-59000 \, \text{J/mol}}{2 \times 96500 \, \text{C/mol}} \]
\[ E = \frac{59000}{193000} \approx 0.305 \, \text{V} \]

### Step 4: Calculate the standard cell potential
Now we need to calculate the standard cell potential (\( E^\circ_{\text{cell}} \)):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]

From the problem:
- \( E^\circ_{\text{reduction}} = -0.83 \, \text{V} \)
- \( E^\circ_{\text{oxidation}} = -1.23 \, \text{V} \)

Thus:
\[ E^\circ_{\text{cell}} = 0.305 + (-0.83) = -0.525 \, \text{V} \]

### Step 5: Apply the Nernst equation
The Nernst equation at pH = 7 is given by:
\[ E = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]

At 298 K, \( R = 8.314 \, \text{J/(mol K)} \) and \( T = 298 \, \text{K} \):
\[ E = -0.525 - \frac{(8.314)(298)}{2 \times 96500} \ln Q \]

Since \( pH = 7 \), the concentration of \( \text{H}^+ \) is \( 10^{-7} \, \text{M} \).

### Step 6: Calculate the reaction quotient \( Q \)
For the reaction:
\[ Q = \frac{[\text{HCO}_3^-]}{[\text{CO}_3^{2-}][\text{H}^+]^2} \]

Assuming standard conditions and concentrations, we can simplify \( Q \) to:
\[ Q \approx \frac{1}{10^{-14}} = 10^{14} \]

### Step 7: Substitute \( Q \) into the Nernst equation
Now we substitute \( Q \) into the Nernst equation:
\[ E = -0.525 - \frac{(8.314)(298)}{2 \times 96500} \ln(10^{14}) \]
\[ E = -0.525 - \frac{(8.314)(298)}{193000} \times 14 \]

Calculating the values:
\[ E \approx -0.525 - 0.0257 \times 14 \]
\[ E \approx -0.525 - 0.3598 \]
\[ E \approx -0.8848 \, \text{V} \]

### Final Answer
Thus, the potential of the cell at pH = 7 is approximately:
\[ E \approx -0.88 \, \text{V} \]