Question

31. For the cell, \( Hg , Hg _{2} Cl _{2} / Cl l^{-}(0.1 M ) / / Cl ^{-} \) \( (0.01 M ) / Cl _{2}, Pt _{\text {cell }}^{0} \) is \( 1.10 V \). Hence, \( E _{\text {cell }} \) is 1) \( 1.1591 V \) 2) \( -1.1591 V \) 3) \( 1.0409 V \) 4) \( -1.0409 V \)

Answer 1

2Cl^-+2Hg^____>Hg₂Cl₂+2e^-

^2e-+Cl₂^_____>2Cl^-

2Hg+Cl₂^___>Hg₂Cl₂                  E°=1.1V

E=1.1-0.0591/2 *log[cl-]²/[cl^-]^2

Answer 2

\(2Cl^- + 2Hg\rightarrow Hg_2Cl_2+2e^-\)