For the cell shown, the measured cell potential

For the cell shown, the measured cell potential, ?cell, is −0.3553 V at 25 °C.   Pt(s) | H2(g,0.803 atm) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, ?o, are  

For the cell shown, the measured cell potential, ?cell, is −0.3553 V at 25 °C.

 Pt(s) | H2(g,0.803 atm) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s)

The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, ?o, are

 2H+(aq)+2e−⟶H2(g) ?o=0.00 V 

 

 Cd2+(aq)+2e−⟶Cd(s)?o=−0.403 V

 

Calculate the H+ concentration.

The Correct Answer and Explanation is :

To calculate the concentration of ( H^+ ) in the given electrochemical cell, we can use the Nernst equation, which relates the cell potential to the concentrations of the species involved.

Given:

• ( E_{\text{cell}} = -0.3553 \, \text{V} )
• Standard reduction potential for the hydrogen half-cell ( E^\circ_{\text{H}^+/H_2} = 0.00 \, \text{V} )
• Standard reduction potential for the cadmium half-cell ( E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.403 \, \text{V} )
• Partial pressure of ( H_2 ) is 0.803 atm.
• Concentration of ( \text{Cd}^{2+} ) is 1.00 M.
• Temperature ( T = 25^\circ C = 298 \, \text{K} ).

Nernst equation:

The Nernst equation is:

[
E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0592}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right)
]

Where:

• ( E_{\text{cell}} ) is the cell potential.
• ( E^\circ_{\text{cell}} ) is the standard cell potential.
• ( n ) is the number of electrons transferred (which is 2 in this case).
• The ratio involves the concentrations or pressures of the species involved in the reaction.

The overall cell reaction is:

[
\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s) \quad E^\circ = -0.403 \, \text{V}
]

[
2\text{H}^+(aq) + 2e^- \rightarrow H_2(g) \quad E^\circ = 0.00 \, \text{V}
]

Standard Cell Potential:

[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} = 0.00 \, \text{V} – (-0.403 \, \text{V}) = 0.403 \, \text{V}
]

Using the Nernst equation:

[
E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0592}{n} \log \left( \frac{[\text{Cd}^{2+}]}{[\text{H}^+]^2 \cdot P_{\text{H}_2}} \right)
]

Substitute the values into the equation:

[
-0.3553 \, \text{V} = 0.403 \, \text{V} – \frac{0.0592}{2} \log \left( \frac{1.00}{[\text{H}^+]^2 \cdot 0.803} \right)
]

Solve for the concentration of ( H^+ ):

1. Simplify the equation:
   [
   -0.3553 = 0.403 – 0.0296 \log \left( \frac{1.00}{[\text{H}^+]^2 \cdot 0.803} \right)
   ]
2. Rearrange to isolate the logarithmic term:
   [
   -0.7583 = -0.0296 \log \left( \frac{1.00}{[\text{H}^+]^2 \cdot 0.803} \right)
   ]
   [
   \log \left( \frac{1.00}{[\text{H}^+]^2 \cdot 0.803} \right) = 25.64
   ]
3. Solve for ( [H^+] ):
   [
   \frac{1.00}{[\text{H}^+]^2 \cdot 0.803} = 10^{25.64}
   ]
   [
   [\text{H}^+]^2 = \frac{1.00}{0.803 \cdot 10^{25.64}}
   ]
   [
   [\text{H}^+] = 1.21 \times 10^{-13} \, \text{M}
   ]

Thus, the concentration of ( H^+ ) is ( 1.21 \times 10^{-13} \, \text{M} ).

Explanation:

The Nernst equation allows us to calculate the concentration of ions in a solution given the cell potential, standard reduction potentials, and other known factors. By rearranging the equation to solve for ( [H^+] ), we were able to determine its concentration in the solution. The process involves considering both the concentration of cadmium ions and the partial pressure of hydrogen gas, along with the temperature and the cell potential, to find the unknown concentration of hydrogen ions.

Let me generate a visual representation of the Nernst equation for you.

Here is a diagram illustrating the Nernst equation applied to your electrochemical cell. It shows how the different components of the cell, such as the cadmium and hydrogen half-cells, relate to the equation. This visual may help clarify how the standard reduction potentials and concentrations interact in the Nernst equation. Let me know if you’d like to dive deeper into any part of it!