JEE Main 2020 (Online) 7th January Evening Slot | Electrochemistry Question 118 | Chemistry

$${\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$

$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$

$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}$$

$${\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}$$

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MCQ (Single Correct Answer)

-1

Given that the standard potentials (E^o) of Cu²⁺/Cu and Cu⁺/Cu are 0.34 V and 0.522 V respectively, the E^o of Cu²⁺/Cu⁺ :

- 0.182 V

- 0.158 V

0.182 V

+0.158 V

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MCQ (Single Correct Answer)

-1

Given

CO³⁺ + e^– $$ \to $$ CO₂⁺ ; E^o = + 1.81 V

Pb⁴⁺ + 2e^– $$ \to $$ Pb₂⁺ ; E^o = + 1.67 V

Ce⁴⁺ + e^– $$ \to $$ Ce³⁺ ; E^o = + 1.61 V

Bi³⁺ + 3e^– $$ \to $$ Bi ; E^o = + 0.20 V

Oxidizing power of the species will increase in the order :

Co³⁺ < Ce⁴⁺ < Bi³⁺ < Pb⁴⁺

Co³⁺ < Pb⁴⁺ < Ce⁴⁺ < Bi³⁺

Ce⁴⁺ < Pb⁴⁺ < Bi³⁺ < Co³⁺

Bi³⁺ < Ce⁴⁺ < Pb⁴⁺ < Co³⁺

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MCQ (Single Correct Answer)

-1

Which one of the following graphs between molar conductivity ($${\Lambda _m}$$) versus $$\sqrt C $$ is correct ?