1
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 ]
A
Silver and gold in equal mass proportion
B
Silver and gold in proportion to their atomic weights
C
Only gold
D
Only silver
2
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Electrochemistry Question 95 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 :
A
If Eext < 1.1 V, Zn dissolves at anode and Cu deposits at cathode
B
If Eext = 1.1 V, no flow of e or current occurs
C
If Eext > 1.1 V, e flows from Cu to Zn
D
If Eext > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
3
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let CNaCl and CBaSO4 be the conductances (in S) measured for saturated aqueous solutions of NaCl and BaSO4, respectively, at a temperature T. Which of the following is false?
A
Ionic mobilities of ions from both salts increase with T.
B
CNaCl(T2) > CNaCl(T1) for T2 > T1
C
CBaSO4(T2) > CBaSO4(T1) for T2 > T1
D
CNaCl >> CBaSO4 at a given T
4
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The equation that is incorrect is :
A
$${\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$
B
$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$
C
$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}$$
D
$${\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}$$
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