JEE Main 2020 (Online) 5th September Evening Slot | Electrochemistry Question 64 | Chemistry

JEE Main 2020 (Online) 5th September Evening Slot

MCQ (Single Correct Answer)

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The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure. JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Electrochemistry Question 65 English

JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Electrochemistry Question 65 English

The electrolyte X is :

HCl

CH₃COOH

NaCl

KNO₃

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

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250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO₃ and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be

[ $$E_{A{g^ + }/Ag}^0$$ = 0.80 V, $$E_{A{u^ + }/Au}^0$$ = 1.69 V ]

Silver and gold in equal mass proportion

Silver and gold in proportion to their atomic weights

Only gold

Only silver

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

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JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Electrochemistry Question 68 English

$$E_{C{u^{2 + }}|Cu}^0$$ = +0.34 V

$$E_{Z{n^{2 + }}|Zn}^0$$ = -0.76 V

Identify the incorrect statement from the option below for the above cell :

If E_ext < 1.1 V, Zn dissolves at anode and Cu deposits at cathode

If E_ext = 1.1 V, no flow of e^– or current occurs

If E_ext > 1.1 V, e^– flows from Cu to Zn

If E_ext > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode

JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

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The equation that is incorrect is :

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

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

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

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

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