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Electrochemistry question

2019 · 9 Apr · Shift 1 · Q5
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  5. /2019 · 9 Apr · Shift 1 · Q5

Electrochemistry question

2019 · 9 Apr · Shift 1 · Q5

JEE MainChemistryElectrochemistryMCQ+4 / −1
The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is : ZnZnZn(s) + Cu2+Cu^{2+}Cu2+ (aq) →\to→ Zn2+Zn^{2+}Zn2+ (aq) + CuCuCu (s), E° = 2 V at 298 K (Faraday's constant, F = 96000 C mol–1)
  1. A
    384
  2. B
    –192
  3. C
    –384
  4. D
    192
View written solutionFree

Correct answer: C

  1. Use the relation between standard Gibbs energy and cell emf

For an electrochemical cell,

ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circΔG∘=−nFE∘

where:

  • nnn = number of electrons transferred
  • F=96000 C mol−1F = 96000\ \text{C mol}^{-1}F=96000 C mol−1
  • E∘=2 VE^\circ = 2\ \text{V}E∘=2 V
  1. Find the number of electrons transferred

The reaction is:

Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

Half-reactions:

Zn→Zn2++2e−Zn \to Zn^{2+} + 2e^-Zn→Zn2++2e− Cu2++2e−→CuCu^{2+} + 2e^- \to CuCu2++2e−→Cu

So,

n=2n = 2n=2

  1. Substitute into the formula

ΔG∘=−(2)(96000)(2)\Delta G^\circ = -(2)(96000)(2)ΔG∘=−(2)(96000)(2)

ΔG∘=−384000 J mol−1\Delta G^\circ = -384000\ \text{J mol}^{-1}ΔG∘=−384000 J mol−1

  1. Convert to kJ mol−1^{-1}−1

−384000 J mol−1=−384 kJ mol−1-384000\ \text{J mol}^{-1} = -384\ \text{kJ mol}^{-1}−384000 J mol−1=−384 kJ mol−1

  1. Match with the options

Thus,

ΔG∘=−384 kJ mol−1\boxed{\Delta G^\circ = -384\ \text{kJ mol}^{-1}}ΔG∘=−384 kJ mol−1​

So the correct option is C.

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