. Given, E _ Cr ³⁺ / Cr ^0=-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

Step-by-step solution

aligned E _ cell ^0 & = E _ cathode - E _ anode & =-0.42-(-0.72)=0.3 aligned At anode [ Cr Cr ³⁺+3 e ⁻ ] 2 At cathode [ Fe ²⁺+2 e ⁻ Fe ] 32 Cr +3 Fe ²⁺ 2 Cr ³⁺+3 Fe aligned & E = E _ cell ^0- 0.0591 6 [ Cr ³⁺ ]^2 [ Fe ²⁺ ]^3 & =0.3- 0.0591 6 (0.1)^2 (0.01)^3 & =0.26 ~V aligned