Solution

Sn⁴⁺ + 2e^– ⟶ Sn²⁺ E° = 0.13V
Br₂ + 2e^– ⟶ 2Br^– E° = 1.08V Calculate K_eq for the cell formed by two electrodes

10⁴¹
10³²
10^-32
10^-42

Solution

Given :
2Br^– ⟶ Br₂ + 2e^– E = -1.09V;
I2 + 2e ⟶ 2I^– E = 0.54V
Fe²⁺ + 2e ⟶ Fe E = 0.44V
Which of the following reactions will not be spontaneous?

Fe + Br₂ ⟶ FeBr₂
Fe + I₂ ⟶ FeI₂
I₂ + 2Br^- ⟶ 2I^- + Br₂
Br₂ + 2I^- ⟶ 2Br^- + I₂

Solution

In electrolysis of NaCl when Pt electrode is taken then H₂ is liberated at cathode while with Hg cathode it forms sodium amalgam. This is because

Hg is more inert than Pt
more voltage is required to reduce H+ at Hg than at Pt
Na is dissolved in Hg while it does not dissolve in Pt
conc. of H+ions is larger when Pt electrode is taken

Solution

In electrolysis of NaCl when Pt electrode is taken then H₂ liberated at cathode while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H⁺ at Hg than Pt.

On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl₂, all copper of the solution was deposited at cathode. The strength of CuCl₂ solution was(Molar mass of Cu= 63.5; Faraday constant =96,500 C mol^-1)

0.01 N
0.01 M
0.02 M
0.2 N

Solution

The electric charge for electrode decomposition of one gram equivalent of a substance is

one ampere per second
96500 coulombs per second
one ampere for one hour
charge on one mole of electrons

Solution

Charge on one mole of electrons = 96500 C

A galvanic cell is set up from electrodes A and B
Electrode A: \(Cr_{2}O_{7}^{-2}\)/Cr³⁺, E°_red = + 1.33V
Electrode B: Fe³⁺/Fe²⁺, E°_red = 0.77 V
Which of the following statements is false ?

Standard e.m.f of the cell is 0.56 V
Current will flow from electrode A to B in the external circuit
A will act as cathode and have positive polarity
None of these

What is the amount of chlorine evolved when 2 amperes of current is passed for 30 minutes in an aqueous solution of NaCl ?

66 g
1.32 g
33 g
99 g

Solution

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 ?. The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.02M of the same solution is 520? . The molar conductivity of 0.02 M solution of electrolyte will be

1.24 × 10^-4S m² mol^-1/sup>
12.4 × 10^-4S m² mol^-1
124 × 10^-4S m²mol^-1
1240 × 10^-4 S m² mol^-1

Solution

Molar conductances of BaCl₂, H₂SO₄ and HCl at infinite dilutions are x₁, x₂ and x₃ respectively. Equivalent conductance of BaSO₄ at infinite dilution will be :

(x₁ + x₂ – x₃) /2
x₁ + x₂ – 2x₃
(x₁ – x₂ – x₃) / 2
(x₁ + x₂ – 2x₃) /2

Solution

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