For the cell reaction Pb + Sn ²⁺ Pb ²⁺+ Sn Given that, Pb Pb ²⁺, E^ =0.13 ~V Sn ²⁺+2 e⁻ Sn ; E^ =-0.14 ~V What would be the ratio of cation concentration for which E=O ?

Step-by-step solution

For the given cell reaction, E_ cell ^ =-0.14 ~V +0.13 ~V =-0.01 ~V According to Nerst equation, aligned E_ cell & =E_ cell ^ - 0.0591 2 [ Pb ²⁺ ] [ Sn ²⁺ ] 0 & =-0.01= 0.0591 2 [ Pb ²⁺ ] [ Sn ²⁺ ] aligned aligned & [ Pb ²⁺ ] [ Sn ²⁺ ] = 0.01 -0.0296 =-0.3 & [ Pb ²⁺ ] [ Sn ²⁺ ] = antilog (-0.3) & [ Pb ²⁺ ] [ Sn ²⁺ ] =0.5 & aligned