JEE Main2008ChemistryElectrochemistryActual
Given E _ Cr ^3+/ Cr ^ =-0.72 ~V , E _ Fe ^ 2+/ / Fe ^ =-0.42 ~V . The potential for the cell Cr | Cr ³⁺(0.1 M ) | | Fe ²⁺(0.01 M ) | Fe is
Options
- A0.26 ~V
- B0.399 ~V
- C-0.339 ~V
- D-0.26 ~V
Correct answer
A. 0.26 ~V
Step-by-step solution
aligned & As E _ Cr / cr ³⁺ ^0=-0.72 ~V and E _ Fe ²⁺ / Fe ^0=-0.42 ~V & 2 Cr +3 Fe ²⁺ 3 Fe +2 Cr ³⁺ & E _ cell = E _ cell ^0- 0.0591 6 ( Cr ³⁺ )^2 ( Fe ²⁺ )^3 & =(-0.42+0.72)- 0.0591 6 (0.1)^2 (0.01)^3 =0.30- 0.0591 6 (0.1)^2 (0.01)^3 & =0.30- 0.0591 6 10⁻² 10⁻⁶ =0.30- 0.0591 6 10^4 & E _ cell =0.2606 ~V aligned