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

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 :

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

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

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

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

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

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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

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

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The decreasing order of electrical conductivity of the following aqueous solutions is :
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C)

A > B > C

C > B > A

A > C > B

C > A > B

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