JEE Main 20207 Jan, Evening shiftChemistryEasyRedox Reactions and Electrochemistry138× since 2002The equation that is incorrect is :AOption A. (Λm0)KCl−(Λm0)NaCl=(Λm0)KBr−(Λm0)NaBr{\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}(Λm0)KCl−(Λm0)NaCl=(Λm0)KBr−(Λm0)NaBrBOption B. (Λm0)NaBr−(Λm0)NaI=(Λm0)KBr−(Λm0)NaBr{\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}(Λm0)NaBr−(Λm0)NaI=(Λm0)KBr−(Λm0)NaBrCOption C. (Λm0)NaBr−(Λm0)NaCl=(Λm0)KBr−(Λm0)KCl{\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}(Λm0)NaBr−(Λm0)NaCl=(Λm0)KBr−(Λm0)KClDOption D. (Λm0)H2O=(Λm0)HCl+(Λm0)NaOH−(Λm0)NaCl{\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}(Λm0)H2O=(Λm0)HCl+(Λm0)NaOH−(Λm0)NaClCheckSkip · show solution
