JEE Advanced 2012 Paper 2 Chemistry Q11

Chemistry · Chemical Equilibria for Analysis · Electrochemistry · Free Energy & Chemical Equilibrium

Problem

The solubility product (

K_{sp}

\mathrm{mol^3\,dm^{-9}}

) of

\mathrm{MX_2}

298\,\mathrm K

based on the information available for the given concentration cell is (take

2.303\times R\times298/F=0.059\,\mathrm V

)