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`
What is the standard potential Couple of `HCO_(3)^(-)//CO_(3)^(-2),H_(2)`.

Options

A. `-0.305` volts
B. `-0.61` volts
C. `+0.305` volts
D. `+0.61` volts

Answer

To find the standard potential couple of \( \text{HCO}_3^- // \text{CO}_3^{2-}, \text{H}_2 \), we will follow these steps:

### Step 1: Write the Reaction
The equilibrium reaction we are considering is:
\[ \text{HCO}_3^- \rightleftharpoons \text{CO}_3^{2-} + \text{H}^+ \]
This indicates that bicarbonate (\( \text{HCO}_3^- \)) is converting to carbonate (\( \text{CO}_3^{2-} \)) while releasing a proton (\( \text{H}^+ \)).

### Step 2: Calculate the Change in Gibbs Free Energy (\( \Delta G \))
The change in Gibbs free energy for the reaction can be calculated using the formula:
\[ \Delta G = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)} \]
From the problem, we have:
- \( \Delta G_f^\circ (\text{CO}_3^{2-}) = -527.8 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{HCO}_3^-) = -586.8 \, \text{kJ/mol} \)

Substituting these values into the equation:
\[ \Delta G = [-527.8] - [-586.8] = -527.8 + 586.8 = 59 \, \text{kJ/mol} \]

### Step 3: Convert \( \Delta G \) to Joules
Since we need to use Joules in the next calculation, we convert \( \Delta G \):
\[ \Delta G = 59 \, \text{kJ/mol} = 59 \times 10^3 \, \text{J/mol} \]

### Step 4: Calculate the Standard Electrode Potential (\( E \))
The relationship between Gibbs free energy and electrode potential is given by:
\[ E = -\frac{\Delta G}{nF} \]
Where:
- \( n \) is the number of moles of electrons transferred (1 for this reaction)
- \( F \) is Faraday's constant (\( 96500 \, \text{C/mol} \))

Substituting the values:
\[ E = -\frac{59 \times 10^3}{1 \times 96500} \]
Calculating this gives:
\[ E \approx -0.612 \, \text{V} \]

### Step 5: Final Answer
Thus, the standard potential couple of \( \text{HCO}_3^- // \text{CO}_3^{2-}, \text{H}_2 \) is approximately:
\[ E \approx -0.61 \, \text{V} \]