MHT CET202519 Apr 2025Morning ShiftChemistryElectrochemistryActual
For the cell reaction, Zn _ ( s ) +2 Ag _ ( aq ) ⁺ Zn _ ( aq ) ⁺²+2 Ag _ ( s ) Cell potential is less than E ^ cell by 0.0592 V at 298 K when
Options
- A[ Zn ⁺² ]=1 M and [ Ag ⁺ ]=0.1 M
- B[ Zn ⁺² ]=1 M and [ Ag ⁺ ]=0.01 M
- C[ Zn ⁺² ]=0.1 M and [ Ag ⁺ ]=1 M
- D[ Zn ⁺² ]=0.01 M and [ Ag ⁺ ]=1 M
Correct answer
A. [ Zn ⁺² ]=1 M and [ Ag ⁺ ]=0.1 M
Step-by-step solution
Solution: Given the cell reaction: Zn _ ( s ) + 2 Ag _ ( aq ) ⁺ Zn _ ( aq ) ⁺² + 2 Ag _ ( s ) , the Nernst equation at 298 K is E_ cell = E^ _ cell - 0.0592 n Q . Since n = 2 electrons are transferred and Q = [ Zn ⁺²] [ Ag ⁺]^2 , and E_ cell = E^ _ cell - 0.0592 V , substitution yields: E^ _ cell - 0.0592 = E^ _ cell - 0.0592 2 Q . Simplifying, 0.0592 = 0.0592 2 Q , then Q = 2 and Q = 100 . Evaluating each option: A: Q = 1 (0.1)^2 = 100 B: Q = 1 (0.01)^2 = 10^4 C: Q = 0.1 1^2 = 0.1 D: Q = 0.01 1^2 = 0.01 Only optio