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 :
$\left\{ \text{given} , \frac{2.303RT}{F} = 0.06 V \; at \; 298K\right\}$
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 :
$\left\{ \text{given} , \frac{2.303RT}{F} = 0.06 V \; at \; 298K\right\}$
Updated On: July 22, 2025
- 0.20 V
- 0.76 V
- 0.40 V
- 0.94 V
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The Correct Option is A
Solution and Explanation
${Pt(s) H_2 (g, 1bar) HCl(aq) AgCl(s) Ag(s) |Pt(s) }$
Anode: ${H_2 ->[10^{-6} m] 2H^{+} + 2e \times 1} $
Cathode : ${e^{-} + AgCl(s) -> Ag(s) + Cl^{-} (aq) }$
${H_2 (g) l + AgCl(s) -> 2H^{+} + 2Ag(s) + 2 Cl^{-} (aq)}$
${E_{cell} = E^{0}_{cell} - \frac{0.06}{2} \log_{10} ((H^{+})^{2} . (Cl^{-})^{2})}$
${.925 = (E^{0}_{H_2 /H^{+}}} + E^{0}_{AgCl / Ag, Cl^{-}}) - \frac{0.06}{2} \log_{10} ((10^{-6})^{2} (10^{-6})^{2})}$
$.92 = 0 + E^{0}_{AgCl /Ag, Cl^{-}} - 0.03 \log_{10} (10^{-6} )^4$
${E^{0}_{AgCl } / Ag, Cl^{-} = .9.2 + .03 \times -24 = 0.2 \; V }$
Anode: ${H_2 ->[10^{-6} m] 2H^{+} + 2e \times 1} $
Cathode : ${e^{-} + AgCl(s) -> Ag(s) + Cl^{-} (aq) }$
${H_2 (g) l + AgCl(s) -> 2H^{+} + 2Ag(s) + 2 Cl^{-} (aq)}$
${E_{cell} = E^{0}_{cell} - \frac{0.06}{2} \log_{10} ((H^{+})^{2} . (Cl^{-})^{2})}$
${.925 = (E^{0}_{H_2 /H^{+}}} + E^{0}_{AgCl / Ag, Cl^{-}}) - \frac{0.06}{2} \log_{10} ((10^{-6})^{2} (10^{-6})^{2})}$
$.92 = 0 + E^{0}_{AgCl /Ag, Cl^{-}} - 0.03 \log_{10} (10^{-6} )^4$
${E^{0}_{AgCl } / Ag, Cl^{-} = .9.2 + .03 \times -24 = 0.2 \; V }$
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Concepts Used:
Galvanic Cells
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. It converts chemical energy to electrical energy.
It consists of two half cells and in each half cell, a suitable electrode is immersed. The two half cells are connected through a salt bridge. The need for the salt bridge is to keep the oxidation and reduction processes running simultaneously. Without it, the electrons liberated at the anode would get attracted to the cathode thereby stopping the reaction on the whole.
Working Principle Of Galvanic Cell:
- Take two beakers containing electrolytic solutions of copper sulphate and zinc sulphate are taken. It is connected via a salt bridge containing an aqueous solution of potassium chloride.
- Zinc and copper electrodes are immersed in the respective electrodes and connected through a voltmeter to measure the electrical potential.
- Zinc which acts as the anode readily undergoes an oxidation process and acquires a negative charge.
- The electrons travel through the salt bridge and undergo a reduction process at the copper cathode.
- Thus the cathode would acquire a positive charge.
- This flow of electrons from the anode to the cathode induces a flow of electric current in the opposite direction which shall be measured by the voltmeter.
Types of Voltaic Cell:
- Primary Cell
- Dry Cell
- Mercury Cell
- Alkaline Cell
- Secondary Cell
- Nickel-Cadmium Cell
- Lead-Acid Cell
- Lithium-Ion Cell