The E ^ of M M ²⁺ | Cu ²⁺ / Cu is 0.3 V , At what concentration of Cu ²⁺ (in mol L ⁻¹ ), the E _ cell value becomes zero? ( 2.303 R T F =0.06 ) ( Conc. of M ²⁺=0.1 M )

Step-by-step solution

According to Nernst equation, E = E ^ - 2.303 RT nF [ product ] [ reactant ] E = E ^ - 2.303 RT nF [ M ²⁺ ] [ Cu ²⁺ ] Given, aligned & E ^ =0.3 ~V & 2.303 RT F =0.06 aligned aligned & n =2 & M ²⁺=0.1 M aligned We have to take E _ cell =0 . According to question. 0=0.3 ~V - 0.06 3 ₁₀ [10⁻¹ ] [ Cu ²⁺ ] -0.3=-0.03 ₁₀ [10⁻¹ ] [ Cu ²⁺ ] 10= ₁₀ [10⁻¹ ] [ Cu ²⁺ ] [10⁻¹ ] [ Cu ²⁺ ] =10¹⁰ or, [ Cu ²⁺ ]= 10⁻¹ 10¹⁰ =10⁻¹⁻¹⁰=10⁻¹¹ Cu ²⁺=10⁻¹¹