Limiting molar conductivity of NH 4 OH (i.e., Λ m NH 4 OH 0 is equal to -

Λ m NH 4 Cl 0 + Λ m NaOH 0 - Λ m NaCl 0

Λ m NaOH 0 + Λ m NaCl 0 - Λ m NaOH 0

Λ m NaOH 0 + Λ m NaCl 0 - Λ m NH 4 Cl 0

Λ m NH 4 OH 0 + Λ m NH 4 Cl 0 - Λ m H C l 0

Chemistry - Electrochemistry - Quiz-7

An excess of granular zinc was added to 500 mL of 1M nickel nitrate till equilibrium was established. Find out the concentration of nickel at the equilibrium, if the standard electrode potential of $Z n^{2 +} / Z n a n d N i^{2 +} / N i$ are – 0.75 and – 0.24 V, respectively.

0%
5.56 x 10^-18 M
0%
0.56 x 10^-18 M
0%
2.16 x 10^-18 M
0%
4.12 x 10^-18 M

Calculate the cell EMF in mV for $P t | H_{2} (1 a t m) | H C l (0.01 M) | A g C l (s) | A g (s) a t 298 K$

If $∆ {G_{r}}^{°}$ values are at $25 ° C$ $- 109.56 \frac{k J}{m o l} f o r A g C l (s) & - 130.79 \frac{k J}{m o l} f o r (H^{+} + C l^{-}) (a q)$

0%

456 mV
0%

654 mV
0%

546 mV
0%

None of these

A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of

copper was deposited on the cathode. Calculate the current efficiency for the copper

deposition. (Cu – 63.5)

0%

79%
0%

39.5%
0%

63.25%
0%

63.5%

In acidic medium $M n {O_{4}}^{-}$ is an oxidising agent. $M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O . I f H^{+}$ ion concentration is doubled, electrode potential of the half cell will :

0%

Increase by 28.46 mV
0%

Decrease by 28.46 mV
0%

Increase by 14.23 mV
0%

Decrease by 142.30 mV

The amount of charge required for the oxidation of 1 mol of $H_{2} O_{2} to O_{2}$ is:

0%

$9.65 \times 10^{4} C$
0%

93000 C
0%

$1.93 \times 10^{5} C$
0%

$19.3 \times 10^{2} C$

The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (Atomic mass of Ca=40 g mol^-1) is:

0%

2
0%

3
0%

4
0%

1

On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:

0%

Oxygen gas
0%

$H_{2} S$ gas
0%

${SO}_{2}$ gas
0%

Hydrogen gas

Calculate the EMF of the cell at 298 K $P t | H_{2} (1 a t m) | N a O H (x M), N a C l (x M) | A g C l (s) | A g$ ${E^{°}}_{C l^{-} / A g C l / A g} = + 0.222 V$

0%

1.048 V
0%

emf depends on x and cannot be determined unless value of x is given
0%

– 0.604 V
0%

– 0.04 V

Assertion : Fluorine cannot be prepared from fluorides by chemical oxidation.

Reason : Fluorine is the strongest oxidizing agent due to its highly positive standard potential.

0%
If both the ssertion and the reason are true and the reason is a correct explanation of the assertion
0%
If oth the assertion and reason are true but the reason is not a correct explanation of the assertion
0%
If the assertion is true but the reason is false
0%
If both the assertion an reason are false

Assertion : : Sodium reacts with water at room temperature, whereas magnesium decomposes only hot water.

Reason : Sodium is more electropositive than magnesium.

0%
If both the ssertion and the reason are true and the reason is a correct explanation of the assertion
0%
If oth the assertion and reason are true but the reason is not a correct explanation of the assertion
0%
If the assertion is true but the reason is false
0%
If both the assertion an reason are false

Assertion : Silver nitrate solution becomes bluish when copper rod is placed in it.

Reason : Copper is more electropositive than silver.

0%
If both the ssertion and the reason are true and the reason is a correct explanation of the assertion
0%
If oth the assertion and reason are true but the reason is not a correct explanation of the assertion
0%
If the assertion is true but the reason is false
0%
If both the assertion an reason are false

The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 x 10^-3Scm^-1 at 298 K is -

0%

11.5 S ${cm}^{2} / mol$
0%

21.5 S ${cm}^{2} / mol$
0%

31.5 S ${cm}^{2} / mol$
0%

41.5 S ${cm}^{2} / mol$

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

Then the species undergoing disproportionation is:-

0%
0%
0%
0%

HBrO

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be :

0%

0.59 V
0%
0.118 V
0%
1.18 V
0%
0.059 V

At 25 ºC molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm^–1cm² mol^–1 and at infinite dilution, its molar conductance is 238 ohm^–1 cm² mol^–1. The degree of ionization of ammonium hydroxide at the same concentration and temperature is :

0%

20.800%
0%

4.008%
0%

40.800%
0%

2.080%

A button cell used in watches functions as following

Zn(s) + Ag₂O(s) + H₂O(l) 2Ag(s) + Zn²⁺ (aq) + 2OH^– (aq)

If half cell potentials are :

Zn²⁺ (aq) + 2e^– → Zn(s); E º = – 0.76 V

Ag₂O (s) + H₂O (l) + 2e^– → 2Ag(s) + 2OH^– (aq); E º = 0.34 V

The cell potential will be:

0%

0.42 V
0%

0.84 V
0%

1.34 V
0%

1.10 V

Limiting molar conductivity of NH₄OH (i.e., $Λ_{m ({NH}_{4} OH)}^{0}$ is equal to -

0%

$Λ_{m (NaOH)}^{0} + Λ_{m (NaCl)}^{0} - Λ_{m (NaOH)}^{0}$
0%

$Λ_{m (NaOH)}^{0} + Λ_{m (NaCl)}^{0} - Λ_{m ({NH}_{4} Cl)}^{0}$
0%

$Λ_{m ({NH}_{4} OH)}^{0} + Λ_{m ({NH}_{4} Cl)}^{0} - Λ_{m (H C l)}^{0}$
0%

$Λ_{m ({NH}_{4} Cl)}^{0} + Λ_{m (NaOH)}^{0} - Λ_{m (NaCl)}^{0}$

Standard electrode potential of three metals X, Y and Z are -1.2 V, +0.5 V and -3.0 V respectively. The reducing power of these metals will be :

0%

Y > X > Z
0%

Z > X > Y
0%

X > Y > Z
0%

Y > Z > X

The electrode potentials for

{Cu}^{2 +} (aq) + e^{-} \to {Cu}^{+} (aq) and {Cu}^{+} (aq) + e^{-} \to Cu (s)

are +0.15 V and +0.50 V respectively. The value of

E_{{Cu}^{2 +} / Cu}^{0}

will be :

0%

0.325 V
0%

0.650 V
0%

0.150V
0%

0.500 V

Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is -0.74. These two couple in their standard state are connected to make a cell. The cell potential will be :

0%

+0.89 V
0%

+0.18 V
0%

+1.83 V
0%

+1.199 V

For the reduction of silver ions with copper metal, the standard cell potential was found to be + 0.46 V a t 25 °C. The value of standard Gibbs energy, ΔG^o will be :

(F = 96500C mol^-1)

0%

-89.0 kJ
0%

-89.0 J
0%

-44.5 kJ
0%

-98.0 kJ

The molar conductance of $\frac{M}{32}$ solution of a weak monobasic acid is 8.0 ohm^-1 cm²and at infinite dilution is 400 ohm^-1 cm². The dissociation constant of this acid is :

0%

$1.25 \times 10^{- 5}$
0%

$1.25 \times 10^{- 6}$
0%

$6.25 \times 10^{- 4}$
0%

$1.25 \times 10^{- 4}$

The correct increasing order of reducing power of the metals is :

(K⁺/K = –2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = –2.37 V, Cr³⁺/Cr = – 0.74V)

0%

Cr < Mg < K < Ag < Hg
0%

Mg < K < Ag < Hg < Cr
0%

K < Ag < Hg < Cr < Mg
0%

Ag < Hg < Cr < Mg < K

Given,
(i) ${Cu}^{2 +} + 2 e^{-} \to Cu$ E^o = 0.337 V
(ii) ${Cu}^{2 +} + e^{-} \to {Cu}^{+}$ E^o = 0.153 V
Electrode potential, E⁰ for the reaction,
${Cu}^{+} + e^{-} \to Cu$ , will be :

0%

0.52 V
0%

0.90 V
0%

0.30 V
0%

0.38 V

The correct order of ionic mobility of ions in the aqueous solution is :

0%

Na⁺> K⁺> Rb⁺> Cs⁺
0%

K⁺> Na⁺> Rb⁺> Cs⁺
0%

Cs⁺> Rb⁺> K⁺> Na⁺
0%

Rb⁺> K⁺> Cs⁺> Na⁺

A steady current of 1.5 A flows through a copper voltmeter for 10 min. If the electrochemical equivalent of copper is 30 × 10^-5 gC^-1, the mass of copper deposited on the electrode will be :

0%

0.40 g
0%

0.50 g
0%

0.67 g
0%

0.27 g

The efficiency of a fuel cell is given by :

0%

$\frac{∆ H}{∆ G}$
0%

$\frac{∆ G}{∆ S}$
0%

$\frac{∆ G}{∆ H}$
0%

$\frac{∆ S}{∆ G}$

In producing chlorine through electrolysis 100 W power at 125 V is being consumed. Liberation of chlorine per min is -

(ECE of chlorine is 0.367×10^-6 kg/C)

0%

17.6 mg
0%
21.3 mg
0%
24.3 mg
0%
13.6 mg

If $E_{{Fe}^{2 +} / Fe}^{o}$ = -0.441 V and $E_{{Fe}^{3 +} / {Fe}^{2 +}}^{o}$ = 0.771 V, the standard emf of the reaction :

Fe + 2Fe³⁺→ 3Fe²⁺ will be :

0%

0.330 V
0%

1.653 V
0%

1.212 V
0%

0.111 V

For the cell reaction $2 {Fe}^{3 +} (aq) + 2 I^{-} (aq) \to 2 {Fe}^{2 +} (aq) + I_{2} (aq)$

$E_{cell}^{⊝} = 0.24 V$ at $298 K$ . The standard Gibbs energy $(∆, G^{⊝})$ of the cell reaction is:

[Given: $F = 96500 C {mol}^{- 1}$ ]

0%

$23.16 kJ {mol}^{- 1}$
0%

$- 46.32 kJ {mol}^{- 1}$
0%

$- 23.16 kJ {mol}^{- 1}$
0%

$46.32 kJ {mol}^{- 1}$

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Practice Chemistry Quiz Questions and Answers

0:0:1

Practice Chemistry Quiz Questions and Answers

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