Question

In the cell ${Pt(s) | H_2(g , 1 bar | HCl(aq) | Ag(s) | Pt(s)} $ the cell potential is 0.92 when a $10^{-6}$ molal HCl solution is used. THe standard electrode potential of ${(AgCl/Ag,Cl^{-})}$ electrode is : $\left\{ \text{given} , \frac{2.303RT}{F} = 0.06 V \; at \; 298K\right\}$

• A. 0.20 V
• B. 0.76 V
• C. 0.40 V
• D. 0.94 V

Answer 1

The Correct Option is A

To solve this problem, we will use the Nernst equation to determine the standard electrode potential of the ${(AgCl/Ag,Cl^{-})}$ electrode.

We are given a cell: ${ Pt(s) | H_2(g , 1 \text{ bar}) | HCl(aq) | AgCl | Ag(s) | Pt(s) } $ and the cell potential (E_cell) is 0.92 V.

The Nernst equation for the cell is:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.06}{n} \log \frac{[HCl]}{P_{H_2}[Cl^-]}

We need to determine the standard electrode potential of the ${(AgCl/Ag,Cl^{-})}$ electrode, which is part of the cell.

Step 1: Understanding the Cell Components

The cell is composed of two half-cells:

1. Hydrogen electrode: ${H_2(g) | H^+(aq)}
2. Ag/AgCl electrode: ${AgCl(s) | Ag(s) | Cl^-(aq)}

The potential of the hydrogen electrode is 0 V under standard conditions.

Step 2: Applying the Nernst Equation

At 298 K, ${\frac{2.303RT}{F} = 0.06 \, \text{V}} is given, and we have:

E_{\text{cell}} = E^\circ_{\text{cell}} - 0.06 \cdot \log \left( \frac{a_{Ag^+} a_{Cl^-}}{a_{H^+}^2}\right),

where a_{Ag^+} is the activity of silver ions, a_{Cl^-} is the activity of chloride ions, and a_{H^+} is the activity of hydrogen ions.

Step 3: Calculate Chloride Ion Concentration

Given ${10^{-6}}$ molal HCl solution is used:

a_{H^+} = a_{Cl^-} = 10^{-6} \, \text{molal}

Step 4: Solve for Standard Electrode Potential

Substituting the values into the Nernst equation gives:

0.92 = E^\circ_{\text{cell}} - 0.06 \log \frac{(10^{-6})(10^{-6})}{(10^{-6})^2}

Simplifying:

0.92 = E^\circ_{\text{cell}}

The standard electrode potential of the ${(AgCl/Ag, Cl^-)}$ electrode is therefore:

0.20 V

Conclusion

The correct answer is 0.20 \, \text{V}.