Q35. The cell Zn|Zn²+ (1M)||Cu²+ (1 M) Cu. (E°Cell = 1.10 V) Was allowed to be completely discharged

Q: What is the correct answer for CUET UG Chemistry Q35?

The correct answer is D. The electrochemical cell is Zn(s) | Zn²⁺(1M) || Cu²⁺(1M) | Cu(s).The standard cell potential (E°Cell) = 1.10 V.At complete discharge, the cell potential (Ecell) = 0 V.The Nernst equation for this cell is:Ecell = E°cell - (0.0592 / n) log([Zn²⁺] / [Cu²⁺])Here, n = 2 (number of electrons transferred in the reaction Zn + Cu²⁺ → Zn²⁺ + Cu).Substitute Ecell = 0 and E°cell = 1.10 V:0 = 1.10 - (0.0592 /

Q: Q35. The cell Zn|Zn²+ (1M)||Cu²+ (1 M) Cu. (E°Cell = 1.10 V) Was allowed to be completely discharged at 298 K. The relative concentration of Zn²+/Cu²+ is: (a) 9.65 x 104 (b) Antilog 24.8 (c) 37.7 (d)…

The correct answer is D. The electrochemical cell is Zn(s) | Zn²⁺(1M) || Cu²⁺(1M) | Cu(s).The standard cell potential (E°Cell) = 1.10 V.At complete discharge, the cell potential (Ecell) = 0 V.The Nernst equation for this cell is:Ecell = E°cell - (0.0592 / n) log([Zn²⁺] / [Cu²⁺])Here, n = 2 (number of electrons transferred in the reaction Zn + Cu²⁺ → Zn²⁺ + Cu).Substitute Ecell = 0 and E°cell = 1.10 V:0 = 1.10 - (0.0592 /

Q: What are the options for this CUET UG Chemistry question?

A. (a) 9.65 x 104 · B. (b) Antilog 24.8 · C. (c) 37.7 · D. (d) 1037.3

Question

Q35. The cell Zn|Zn²+ (1M)||Cu²+ (1 M) Cu. (E°Cell = 1.10 V) Was allowed to be completely discharged at 298 K. The relative concentration of Zn²+/Cu²+ is:
(a) 9.65 x 104
(b) Antilog 24.8
(c) 37.7
(d) 1037.3

FirstInTest · Accepted Answer

The electrochemical cell is Zn(s) | Zn²⁺(1M) || Cu²⁺(1M) | Cu(s).The standard cell potential (E°Cell) = 1.10 V.At complete discharge, the cell potential (Ecell) = 0 V.The Nernst equation for this cell is:Ecell = E°cell - (0.0592 / n) log([Zn²⁺] / [Cu²⁺])Here, n = 2 (number of electrons transferred in the reaction Zn + Cu²⁺ → Zn²⁺ + Cu).Substitute Ecell = 0 and E°cell = 1.10 V:0 = 1.10 - (0.0592 / 2) log([Zn²⁺] / [Cu²⁺])(0.0592 / 2) log([Zn²⁺] / [Cu²⁺]) = 1.100.0296 log([Zn²⁺] / [Cu²⁺]) = 1.10log([Zn²⁺] / [Cu²⁺]) = 1.10 / 0.0296 ≈ 37.16[Zn²⁺] / [Cu²⁺] = 10^37.16 ≈ 10^37.3 (rounded).Therefore, the relative concentration of Zn²⁺/Cu²⁺ is approximately 10^37.3.

Answer

(a) 9.65 x 104

Answer

(b) Antilog 24.8

Answer

(c) 37.7

Answer

(d) 1037.3

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