Zn | Zn 2+ (a = 0.1 M) || Fe 2+ (a = 0.01 M) | Fe. The emf of the above cell is 0.2905V…
Chemistry · JEE Advanced · NTA Exams — Electrochemistry
Zn | Zn2+ (a = 0.1 M) || Fe2+ (a = 0.01 M) | Fe.
The emf of the above cell is 0.2905V. Equilibrium constant for the cell reaction is (2004)
The emf of the above cell is 0.2905V. Equilibrium constant for the cell reaction is (2004)
- 100.32/0.059
- 100.32/0.0295
- 100.26/0.0295
- 100.32/0.295
Answer
(B) 10 0.32/0.0295
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- 1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was…
- The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is (2001)
- For the following electrochemical cell at 298 K, Pt ( s ) |H 2( g ) , (1 bar)|H + ( aq ) , (1 M)|| M 4 + ( aq…
- On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl 2 , all copper of…
- The conductance of 0.1 M HCl solution is greater than that of 0.1 M NaCl. This is because
- Conductance (unit Siemen’s S) is directly proportional to area of the vessel and the concentration of the…
- The standard reduction potential values of the three metallic cations X, Y and Z are 0.52, – 3.03 and – 1.18V…
- The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is
More Electrochemistry questions · Browse all practice questions