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Electrochemistry question

2025 · 29 Jan · Shift 1 · Q2
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Electrochemistry question

2025 · 29 Jan · Shift 1 · Q2

JEE MainChemistryElectrochemistryMCQ+4 / −1
For a MgMgMg | Mg2+Mg^{2+}Mg2+ (aq) || Ag+Ag^+Ag+ (aq) | AgAgAg the correct Nernst Equation is :
  1. A
    Ecell =Ecell o−RT2 Fln⁡[Ag+]2[Mg2+]\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}Ecell ​=Ecell o​−2 FRT​ln[Mg2+][Ag+]2​
  2. B
    Ecell =Ecell o−RT2 Fln⁡[Mg2+][Ag+]\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]}Ecell ​=Ecell o​−2 FRT​ln[Ag+][Mg2+]​
  3. C
    Ecell =Ecell o−RT2 Fln⁡[Ag+][Mg2+]\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Mg}^{2+}\right]}Ecell ​=Ecell o​−2 FRT​ln[Mg2+][Ag+]​
  4. D
    Ecell =Ecell o+RT2 Fln⁡[Ag+]2[Mg2+]\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}Ecell ​=Ecell o​+2 FRT​ln[Mg2+][Ag+]2​
View written solutionFree

Correct answer: D

  1. Write the half-reactions for the cell

Given cell: Mg∣Mg2+(aq)∣∣Ag+(aq)∣AgMg|Mg^{2+}(aq)||Ag^+(aq)|AgMg∣Mg2+(aq)∣∣Ag+(aq)∣Ag

  • Anode (oxidation): Mg(s)→Mg2+(aq)+2e−Mg(s) \rightarrow Mg^{2+}(aq) + 2e^-Mg(s)→Mg2+(aq)+2e−

  • Cathode (reduction): Ag+(aq)+e−→Ag(s)Ag^+(aq) + e^- \rightarrow Ag(s)Ag+(aq)+e−→Ag(s)

To balance electrons, multiply the silver half-reaction by 222: 2Ag+(aq)+2e−→2Ag(s)2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)2Ag+(aq)+2e−→2Ag(s)

  1. Overall cell reaction

Adding the two half-reactions: Mg(s)+2Ag+(aq)→Mg2+(aq)+2Ag(s)Mg(s) + 2Ag^+(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)Mg(s)+2Ag+(aq)→Mg2+(aq)+2Ag(s)

  1. Write the reaction quotient QQQ

Solids are omitted from QQQ, so: Q=[Mg2+][Ag+]2Q = \frac{[Mg^{2+}]}{[Ag^+]^2}Q=[Ag+]2[Mg2+]​

  1. Apply the Nernst equation

General form: Ecell=Ecell∘−RTnFln⁡QE_{cell} = E_{cell}^\circ - \frac{RT}{nF}\ln QEcell​=Ecell∘​−nFRT​lnQ

Here, number of electrons transferred is: n=2n=2n=2

So, Ecell=Ecell∘−RT2Fln⁡([Mg2+][Ag+]2)E_{cell} = E_{cell}^\circ - \frac{RT}{2F}\ln \left(\frac{[Mg^{2+}]}{[Ag^+]^2}\right)Ecell​=Ecell∘​−2FRT​ln([Ag+]2[Mg2+]​)

This can also be written as: Ecell=Ecell∘+RT2Fln⁡([Ag+]2[Mg2+])E_{cell} = E_{cell}^\circ + \frac{RT}{2F}\ln \left(\frac{[Ag^+]^2}{[Mg^{2+}] }\right)Ecell​=Ecell∘​+2FRT​ln([Mg2+][Ag+]2​)

  1. Match with the options

This exactly matches Option D: Ecell=Ecello+RT2Fln⁡[Ag+]2[Mg2+]E_{cell}=E_{cell}^{o}+\frac{RT}{2F}\ln \frac{[Ag^+]^2}{[Mg^{2+}]}Ecell​=Ecello​+2FRT​ln[Mg2+][Ag+]2​

Therefore, the correct answer is D.

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