In the cell reaction,
`4Cr^(2+) (aq) +O_(2) (g) +4H^(+) (aq) hArr 4Cr^(3+) (aq) +2H_(2)O(I)`,
the concentrations are :
`[Cr^(2+)]=0.1 M, [Cr^(3+)]=0.082 M, [H^(+)]=0.01M`. Find the partial pressure of `O_(2)` gas at equilibrium at `25^(@)C`.
`[E_(Cr^(3+)//Cr^(2+))^(Theta) =-0.41V.E_(O_(2)//H_(2)O)^(Theta)=1.23V]`.

Explanation

To find the partial pressure of \( O_2 \) gas at equilibrium for the given cell reaction:

\[ 4Cr^{2+} (aq) + O_2 (g) + 4H^+ (aq) \rightleftharpoons 4Cr^{3+} (aq) + 2H_2O (l) \]

with the given concentrations:

- \([Cr^{2+}] = 0.1 \, M\)
- \([Cr^{3+}] = 0.082 \, M\)
- \([H^+] = 0.01 \, M\)

and standard reduction potentials:

- \(E^\circ_{Cr^{3+}/Cr^{2+}} = -0.41 \, V\)
- \(E^\circ_{O_2/H_2O} = 1.23 \, V\)

we can use the Nernst equation to find the partial pressure of \( O_2 \).

### Step 1: Determine the standard cell potential \( E^\circ_{cell} \)

The overall cell reaction involves the oxidation of \( Cr^{2+} \) to \( Cr^{3+} \) and the reduction of \( O_2 \) to \( H_2O \). The standard cell potential can be calculated as follows:

\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Here, the cathode reaction is the reduction of \( O_2 \) and the anode reaction is the oxidation of \( Cr^{2+} \):

\[ E^\circ_{cell} = E^\circ_{O_2/H_2O} - E^\circ_{Cr^{3+}/Cr^{2+}} = 1.23 \, V - (-0.41 \, V) = 1.23 \, V + 0.41 \, V = 1.64 \, V \]

### Step 2: Use the Nernst equation

The Nernst equation is given by:

\[ E = E^\circ - \frac{0.059}{n} \log Q \]

where \( n \) is the number of moles of electrons transferred in the balanced equation, and \( Q \) is the reaction quotient.

### Step 3: Determine \( n \)

From the balanced equation, we see that 4 moles of \( Cr^{2+} \) are oxidized to 4 moles of \( Cr^{3+} \), which means 4 electrons are transferred:

\[ n = 4 \]

### Step 4: Calculate the reaction quotient \( Q \)

The reaction quotient \( Q \) for the reaction is given by:

\[ Q = \frac{[Cr^{3+}]^4}{[Cr^{2+}]^4 \cdot [H^+]^4 \cdot P_{O_2}} \]

Substituting the known concentrations:

\[ Q = \frac{(0.082)^4}{(0.1)^4 \cdot (0.01)^4 \cdot P_{O_2}} \]

Calculating \( (0.082)^4 \):

\[ (0.082)^4 = 0.0000452 \, (approximately) \]

Calculating \( (0.1)^4 \):

\[ (0.1)^4 = 0.0001 \]

Calculating \( (0.01)^4 \):

\[ (0.01)^4 = 0.00000001 \]

Now substituting these values into \( Q \):

\[ Q = \frac{0.0000452}{0.0001 \cdot 0.00000001 \cdot P_{O_2}} = \frac{0.0000452}{0.000000000001 \cdot P_{O_2}} = \frac{0.0000452 \times 10^{9}}{P_{O_2}} = \frac{45200}{P_{O_2}} \]

### Step 5: Substitute \( Q \) into the Nernst equation

At equilibrium, \( E = 0 \):

\[ 0 = 1.64 - \frac{0.059}{4} \log \left( \frac{45200}{P_{O_2}} \right) \]

Rearranging gives:

\[ \frac{0.059}{4} \log \left( \frac{45200}{P_{O_2}} \right) = 1.64 \]

Multiplying both sides by \( \frac{4}{0.059} \):

\[ \log \left( \frac{45200}{P_{O_2}} \right) = \frac{1.64 \times 4}{0.059} \approx 111.18 \]

### Step 6: Solve for \( P_{O_2} \)

Taking the antilog:

\[ \frac{45200}{P_{O_2}} = 10^{111.18} \]

Thus,

\[ P_{O_2} = \frac{45200}{10^{111.18}} \approx 4.52 \times 10^{-104} \, atm \]

### Final Answer

The partial pressure of \( O_2 \) gas at equilibrium is:

\[ P_{O_2} \approx 4.52 \times 10^{-104} \, atm \]