MHT CET202525 Apr 2025Morning ShiftChemistryElectrochemistryActual
If E^ ( Zn _ ( aq ) ⁺² Zn _ ( s ) )=-0 76 ~V . Calculate potential for Zn _ ( s ) Zn _ (0.01 M ) ⁺²+2 e at 298 K ?
Options
- A+0.8192 ~V
- B-0.8192
- C+0.7008 ~V
- D-0.7008 ~V
Correct answer
A. +0.8192 ~V
Step-by-step solution
The standard oxidation potential is the negative of the standard reduction potential: E^ _ ox = -(-0.76 V ) = +0.76 V . The Nernst equation for oxidation at T = 298 K is E = E^ _ ox - 0.0592 n Q , where n = 2 and Q = [ Zn ²⁺] = 0.01 . Substituting these values: E = 0.76 - 0.0592 2 (10⁻²) Since (10⁻²) = -2 , the result is E = 0.76 + 0.0592 = 0.8192 V . Final potential: +0.8192 V