AP EAMCET201920 Apr 2019Morning ShiftChemistryElectrochemistryActual
Under which of the following conditions (E ) value of the cell, for the cell reaction given is maximum? ( Zn (s)+ Cu ²⁺(a q) Cu (s)+ Zn ²⁺(a q) ) ( ( array l 2.303 R T F at 298 ~K =0.059 ~V , E_ Zn ²⁺ / Zn ^ =-0.76 ~V , E_ Cu ^ 2⁺ / Cu ^0=+0.34 ~V array ) )
Options
- A(C₁=0.1 M , C₂=0.01 M )
- B(C₁=0.01 M , C₂=0.1 M )
- C(C₁=0.1 M , C₂=0.2 M )
- D(C₁=0.2 M , C₂=0.1 M )
Correct answer
A. (C₁=0.1 M , C₂=0.01 M )
Step-by-step solution
From Nernst equation, ( aligned E & =E^ - 2.303 R T n F Q E & =E^ - 2.303 R T n F ( Zn ²⁺ Cu ²⁺ ) E^ _ cell & =E^ _C-E_A^ =0.34-(-0.76) V =1.1 ~V E & =1.1- 0.059 n C₂ C₁ ( array l Zn ²⁺=C₂ Cu ²⁺=C₁ array ) aligned ) By analysing from the above equation, (E ) value of the cell will be maximum when, ( C₂ C₁ ) would come out to be minimum, when ( C₂ C₁ ) value would be minimum then, ( 0.01 0.1 = 10⁻¹ ) (minimum). Thus, option (1) is correct.