NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
The cell, Z n / Z n 2 + ( 1 M ) | | C u 2 + ( 1 M ) / C u ( E c e l l 0 = 1.10 V ) was allowed to be completely discharged at 298K. The relative concentration of Z n 2 + to C u 2 + Z n 2 + C u 2 + is 1 0 x . The value of x is: ( T a k e 2.303 R T F = 0.059 Round off your answer up to one decimal)
Correct answer
37
Step-by-step solution
E c e l l = 0 ; when cell is completely discharged E c e l l = E c e l l 0 - 0.059 2 l o g [ Z n 2 + ] [ C u 2 + ] 0 = 1.1 - 0.059 2 l o g [ Z n 2 + ] [ C u 2 + ] l o g [ Z n 2 + ] [ C u 2 + ] = 2 × 1.1 0.059 = 37.3 ` Z n 2 + C u 2 + = 1 0 37.3 = 1 0 x ∴ x = 37.3