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Electrochemistry question

2024 · 8 Apr · Shift 2 · Q9
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  5. /2024 · 8 Apr · Shift 2 · Q9

Electrochemistry question

2024 · 8 Apr · Shift 2 · Q9

JEE MainChemistryElectrochemistryMCQ+4 / −1
The reaction; 12H2( g)+AgCl(s)→H(aq)++Cl(aq)−+Ag(s)\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}+\mathrm{AgCl}_{(\mathrm{s})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}+\mathrm{Ag}_{(\mathrm{s})}21​H2( g)​+AgCl(s)​→H(aq)+​+Cl(aq)−​+Ag(s)​ occurs in which of the following galvanic cell :
  1. A
    Pt∣H2( g)∣HCl(soln.) ∣AgNO3(aq)∣Ag\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text {soln.) }}\left|\mathrm{AgNO}_{3(\mathrm{aq})}\right| \mathrm{Ag}Pt​H2( g)​​HCl(soln.) ​​AgNO3(aq)​​Ag
  2. B
    Ag∣AgCl(s)∣KCl(soln.) ∣AgNO3 (aq.) ∣Ag\mathrm{Ag}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{KCl}_{\text {(soln.) }}\left|\mathrm{AgNO}_{3 \text { (aq.) }}\right| \mathrm{Ag}Ag​AgCl(s)​​KCl(soln.) ​​AgNO3 (aq.) ​​Ag
  3. C
    Pt∣H2( g)∣KCl(soln.) ∣AgCl(s)∣Ag\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{KCl}_{(\text {soln.) }}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}Pt​H2( g)​​KCl(soln.) ​​AgCl(s)​​Ag
  4. D
    Pt∣H2( g)∣HCl(soln. )∣AgCl(s)∣Ag\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text {soln. })}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}Pt​H2( g)​​HCl(soln. )​​AgCl(s)​​Ag
View written solutionFree

Correct answer: D

  1. Given overall reaction

We need the galvanic cell in which the net reaction is

12H2(g)+AgCl(s)→H(aq)++Cl(aq)−+Ag(s)\frac{1}{2}H_{2(g)} + AgCl_{(s)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} + Ag_{(s)}21​H2(g)​+AgCl(s)​→H(aq)+​+Cl(aq)−​+Ag(s)​

So we identify the two half-reactions.

  1. Write the oxidation and reduction half-reactions

Oxidation half-reaction

Hydrogen is getting oxidized:

12H2→H++e−\frac{1}{2}H_2 \rightarrow H^+ + e^-21​H2​→H++e−

This requires a hydrogen electrode, i.e.

Pt∣H2(g)∣H+(aq)\text{Pt} | H_2(g) | H^+(aq)Pt∣H2​(g)∣H+(aq)

Reduction half-reaction

Silver chloride is getting reduced:

AgCl(s)+e−→Ag(s)+Cl(aq)−AgCl_{(s)} + e^- \rightarrow Ag_{(s)} + Cl^-_{(aq)}AgCl(s)​+e−→Ag(s)​+Cl(aq)−​

This requires a silver-silver chloride electrode, i.e.

Ag∣AgCl(s)∣Cl−(aq)Ag | AgCl_{(s)} | Cl^-(aq)Ag∣AgCl(s)​∣Cl−(aq)

or equivalently written in reverse order as

Cl−(aq)∣AgCl(s)∣AgCl^-(aq) | AgCl_{(s)} | AgCl−(aq)∣AgCl(s)​∣Ag
  1. Combine the required electrodes

Thus the cell must contain:

  • a hydrogen electrode with H+H^+H+ present, so the solution should be HCl, not KCl,
  • a silver-silver chloride electrode with AgCl(s)AgCl(s)AgCl(s) and Cl−Cl^-Cl− present.

So the suitable cell is:

Pt∣H2(g)∣HCl(aq)∣AgCl(s)∣Ag\text{Pt} | H_2(g) | HCl(aq) | AgCl_{(s)} | AgPt∣H2​(g)∣HCl(aq)∣AgCl(s)​∣Ag

which is exactly Option D.

  1. Check each option systematically

Option A

Pt∣H2∣HCl∣AgNO3∣Ag\mathrm{Pt}|H_2|HCl|AgNO_3|AgPt∣H2​∣HCl∣AgNO3​∣Ag

Right electrode is Ag+/AgAg^+/AgAg+/Ag, whose reduction is

Ag++e−→AgAg^+ + e^- \rightarrow AgAg++e−→Ag

This does not involve AgCl(s)AgCl(s)AgCl(s), so it cannot give the required overall reaction.

❌ A is incorrect

Option B

Ag∣AgCl∣KCl∣AgNO3∣AgAg|AgCl|KCl|AgNO_3|AgAg∣AgCl∣KCl∣AgNO3​∣Ag

There is no hydrogen electrode, so the term 12H2→H+\frac12 H_2 \rightarrow H^+21​H2​→H+ cannot occur.

❌ B is incorrect

Option C

Pt∣H2∣KCl∣AgCl∣Ag\mathrm{Pt}|H_2|KCl|AgCl|AgPt∣H2​∣KCl∣AgCl∣Ag

Hydrogen electrode needs H+H^+H+ ions for the half-reaction

12H2→H++e−\frac12 H_2 \rightarrow H^+ + e^-21​H2​→H++e−

But KCl solution does not provide H+H^+H+. So this cell cannot represent the required reaction.

❌ C is incorrect

Option D

Pt∣H2∣HCl∣AgCl∣Ag\mathrm{Pt}|H_2|HCl|AgCl|AgPt∣H2​∣HCl∣AgCl∣Ag

Left electrode:

12H2→H++e−\frac12 H_2 \rightarrow H^+ + e^-21​H2​→H++e−

Right electrode:

AgCl+e−→Ag+Cl−AgCl + e^- \rightarrow Ag + Cl^-AgCl+e−→Ag+Cl−

Adding:

12H2+AgCl→H++Cl−+Ag\frac12 H_2 + AgCl \rightarrow H^+ + Cl^- + Ag21​H2​+AgCl→H++Cl−+Ag

This exactly matches the given reaction.

✅ D is correct

  1. Final answer

The required galvanic cell is:

Pt∣H2(g)∣HCl(aq)∣AgCl(s)∣Ag\boxed{\mathrm{Pt}|H_{2(g)}|HCl_{(aq)}|AgCl_{(s)}|Ag}Pt∣H2(g)​∣HCl(aq)​∣AgCl(s)​∣Ag​

So the correct option is D.

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