Blog Log in Get Started Free

Electrochemistry MCQs for NEET — Chemistry Questions with Answers

Practice free Electrochemistry (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

In group - I , 2 and 3 electrolytc and products at anode and cathode arc mentioned respectively match them appropriately.

Group -1

(1) NaCl (molten)

(2) NaCl (conc. aqueous) (

3) NaCl (dilute aqeous)

(4) $ Al_20_3 , ( +Na_3AIF_6) $

(5) $ KIIF_2 anhydrous HF $

Group -2

(A) $ O_{2(g)} $

(B) $ O_{2(g)} , CO_{(2(g)} $

(D) $ F_{2(g)} $

(E) $ H_{2(g)} $

Group-3

(P) Al metal

(Q) $ Cl_{2(g)} $

(R) Na metal

(S) $ H_{2(g)} and in solution NaOH$

(T ) $ H_{2(g)} $

(1) -(.C) - (.R),

(2) - (.C) - (T) ,

(3)- (A) - (T),

(4) - ( B) - (P) ,

(5) - (D) - ( T)

(1) -(.C) - (.R), (2) - (.C) - (S) , (3)- (A) - (T),. (4) - ( B) - (P) ,. (5) - (D) - ( T)

(1) -(.C) - (.R), (2) - (.C) - (S) , (3)- (A) - (Q), (4) - ( B) - (P) , (5) - (D) - ( T)

(1) -(.C) - (.R), (2) - (.C) - (S) , (3)- (C) - (T), (4) - ( B) - (P) , (5) - (D) - ( T)

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The correct match is: (1) - (C) - (R), (2) - (C) - (S), (3) - (A) - (T), (4) - (B) - (P), (5) - (D) - (T). This matches the electrolytes (molten/aqueous NaCl, Al2O3, KHF2) with the respective products formed at the anode and cathode during electrolysis.

● ● ● ● ●

Potential of a Std. half cell is measured by potentiometer connecting it with S .11.E. here S.H.E. act as anode then potential of half cell would be equals to

A. $ E ^ 0 _{ox} $ of the other electrode B. $ E ^ o_{red } $ of the other electrode C. $ E_{ox} $ of the other electrode D. $ E_{red} $ of the other electrode

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

will be the decrease in the concentration Of $ Ni ^ {2+}_{(aq)} When the reaction Co_(s) + Ni ^ {2+} _{(aq, 0.1 M )} \rightleftharpoons Co^{2+} _{(aq, 0.01 M )} + Ni _{(S)} $ reaches the equilibrium ?

A. 9.78% B. 8.74% C. 95.40% D. 92.50%

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

Which of the following is the strongest oxidising agent [Pb. CET 2000]

$B r O_{3}^{-} / B r^{2 +}, E^{o} = + 1 .50$

$F e^{3 +} / F e^{2 +}, E^{o} = + 0 .76$

$M n O_{4}^{-} / M n^{2 +}, E^{o} = + 1 .52$

$C r_{2} O_{7}^{2 -} / C r^{3 +}, E^{o} = + 1 .33$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Higher is the reduction potential stronger is the oxidising agent. Hence in the given options. $M n O_{4}^{-}$ is strongest oxidising agent.

● ● ● ● ●

An electrochemical cell is shown below Pt, H₂(1 atm)| HCI (0.1 M)CH₃COOH (0.1 M)|

H₂(1 atm), Pt The EMF of the cell will not be zero, because

EMF depends on molarities of acids used

pH of 0.1 M HCl and 0.1 M CH₃COOH is not same

the temperature is constant

acids used in two compartments are different

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) The EMF of the cell will not be zero because concentration of H⁺ ions in two electrolytic solutions is different. Mean HCl is strong acid where, acetic acid is weak acid and gives different pH.

● ● ● ● ●

Saturated solution of KNO₃ is used to make 'salt-bridge' because:

velocity of K⁺ is greater than that of ${NO}_{3}^{-}$

velocity of ${NO}_{3}^{-}$ is greater than that of K⁺

Velocities of both K⁺ and ${NO}_{3}^{-}$ are nearly the same

KNO₃ is highly soluble in water

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

● ● ● ● ●

A current is passed through two voltameters connected in series. The first voltmeter connected in series. The first voltmeter contains XSO₄(aq) while the second voltmeter contains Y₂SO₄(aq). The relative atomic masses of X and Y are in the ratio of 2:1. The ration of the mass of X liberated to the mass of Y liberated is:

1:1

1:2

2:1

none of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) Equal eqivalents of each are liberated.

Eq. of X=Eq. of Y

$\frac{m_{1}}{\frac{2 M}{2}} = \frac{m_{2}}{\frac{M}{1}} ∴ m_{1} = m_{2}$

● ● ● ● ●

The mass of silver(eq. mass = 108) displaced by that quantity of current which displaced 5600 mL of hydrogen at STP is:

54 g

108 g

5.4 g

none of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Concept used: Faradays second law of electrolysis

Solve: Acc. to Faraday's seconf law

$\frac{w_{1}}{e q w t_{1}} = \frac{w_{2}}{e q w t_{2}}$

here since gas is involved se we can directly use

$\frac{w_{a g}}{e q_{a g}} = \frac{v o l . o f g a s a t N T P}{v o l . o f 1 g m e q .} \frac{w_{1}}{108} = \frac{5600}{\frac{22400}{2}} \frac{w_{1}}{108} = \frac{5600}{11200}$

$\frac{w}{108} = \frac{1}{2}, w = \frac{108}{2} = 54 g$

● ● ● ● ●

A silver cup is plated with silver by passing 965 coulomb of electricity. The amount of Ag deposited is:

1.08 g

1.0002 g

9.89 g

107.89 g

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Concept used: Faraday's first law.

Given : Q= 965 coulomb

wt of sub deposited is given by

w = 2 x Q (Faraday's first law)

$w = \frac{E}{96500} \times Q$

Equivalent wt of metal E= $\frac{molecular wt}{n - factor}$

$= w = \frac{\frac{M}{n - factor}}{96500} \times Q = \frac{\frac{108}{1}}{96500} \times 965 = \frac{108}{100} = 1.08 g$

● ● ● ● ●

At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 Ω^-1cm²mol^-1 and at infinite dilution its molar conductance is 238 Ω^-1cm²mol^-1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

2.080 %

20.800 %

4.008 %

40.800 %

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$λ_{c}$ = 9.54Ω^-1cm²mol^-1

Molar conductance at infinite dilution,

${λ_{c}}^{\infty}$ = 238 Ω^-1cm²mol^-1.

We know that,

degree of ionisation,

$α$ = $λ_{c}$ / ${λ_{c}}^{\infty}$

=(9.54/238) x 100 = 4.008%

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Electrochemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.