Electrochemistry Questions

How many coulombs of electricity are consumed when a 100 mA current is passed through a solution of AgNO₃ for half an hour during an electrolysis experiment-

In which one of the following one Faraday of electricity will liberate 1/2 mole of the metal-

The same amount of electricity was passed through two cells containing molten AlCl₃ and molten NaCl. If 1.8 g of Al were liberated in one cell, the amount of Na liberated in the other cell is-

According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte A₂B is-

If x is the specific resistance of the solution and N is the normality of the solution. The equivalent conductivity of the solution is given by-

The molar conductances of HCl, NaCl and CH₃COONa are 426, 126 and 91 Ω⁻¹ cm² mol⁻¹ respectively. The molar conductance for CH₃COOH is-

In the electrolysis of CuCl₂ solution using Cu electrodes the mass of cathode increases by 3.18 g. What happened at the other electrode-

When aqueous solution of KCl is electrolysed, resultant solution-

When an electric current is passed through an aqueous solution of sodium chloride-

What products are formed during the electrolysis of a concentrated aqueous solution of sodium chloride? I : Cl₂(g) II : NaOH(aq) III : H₂(g)

Rust is a mixture of-

During electrolysis of an aqueous solution of copper sulphate, 0.635 g of copper was deposited at cathode. The amount of electricity consumed in coulomb is-

When sodium chloride solution is electrolysed, the gas that is liberated at the cathode is-

On passing electricity through dilute H₂SO₄ solution the amount of substance liberated at the cathode and anode are in the ratio-

When a lead storage battery is charged it acts as-

In electrochemical corrosion of metals, the metal undergoing corrosion -

The Zn acts as sacrificial or cathodic protection to prevent rusting of iron because -

The value of the reaction quotient, Q, for the cell Zn(s)|Zn²⁺(0.01 M)||Ag⁺(1.25 M)|Ag(s) is -

Consider a voltaic cell based on these half-cells Ag⁺ + e⁻ → Ag ; E° = +0.80 V Cd²⁺ + 2e⁻ → Cd ; E° = −0.40 V Identify the anode and give the voltage of the cell under standard conditions -

The cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu is best represented by -

Normal Al–AlCl₃ coupled with standard hydrogen electrode gives an emf of 1.66 V. The standard oxidation electrode potential of aluminium is -

emf of cell Ni | Ni²⁺(1.0 M) || Au³⁺(1.0 M) | Au is..., if E°(Ni²⁺/Ni) = −0.25 V and E°(Au³⁺/Au) = +1.50 V

At equilibrium -

Standard electrode potentials of Fe²⁺ + 2e⁻ → Fe ; E° = −0.440 V Fe³⁺ + 3e⁻ → Fe ; E° = −0.036 V The standard electrode potential (E°) for Fe³⁺ + e⁻ → Fe²⁺ is -

Consider the following equations for a cell reaction A + B ⇌ C + D ; E° = x volt, Keq = K₁ 2A + 2B ⇌ 2C + 2D ; E° = y volt, Keq = K₂ then :

For the reactions MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O ; E° = 1.51 V MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O ; E° = 1.23 V then for the reaction MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O, E° is -

E° values of Mg²⁺|Mg, Fe²⁺|Fe and Zn²⁺|Zn are −2.37 V, −0.44 V and −0.76 V respectively. The correct statement is -

The single electrode potential E of 0.1 M solution of M⁺ ions [E°R = −2.36 V] is -

The standard oxidation potentials are: Zn → Zn²⁺ + 2e⁻ ; E° = 0.76 V Fe → Fe²⁺ + 2e⁻ ; E° = 0.41 V The emf for the reaction Fe²⁺ + Zn → Zn²⁺ + Fe is-

Which of the following will increase the voltage of the cell Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)

Standard E° of half-cell Fe|Fe²⁺ is +0.44 V and standard E° of half-cell Cu|Cu²⁺ is −0.32 V, then-

Consider the following half-cell reactions: I. A + e⁻ → A⁻ E° = +0.96 V II. B⁻ + e⁻ → B²⁻ E° = −0.12 V III. C⁺ + e⁻ → C E° = +0.18 V IV. D²⁺ + 2e⁻ → D E° = −1.12 V What combination of two half-cells would result in a cell with the largest potential?

An aqueous solution of CuSO₄ is stirred with a silver spoon. The following will happen-

The standard electrode potential of Zn, Ag and Cu are −0.76, +0.80 and +0.34 V respectively. Then-

At 298 K, the standard reduction potentials are: Zn²⁺/Zn = −0.762 V Cr³⁺/Cr = −0.740 V 2H⁺/H₂ = 0.00 V Fe³⁺/Fe²⁺ = +0.770 V The strongest reducing agent is-

The reaction ½H₂(g) + AgCl(s) → H⁺(aq) + Cl⁻(aq) + Ag(s) can be represented in the galvanic cell as-

The number of coulombs required to deposit 5.4 g of Al when the electrode reaction is- Al³⁺ + 3e⁻ → Al

On passing 3 ampere of electricity for 50 minutes, 1.8 g of metal deposits. The equivalent mass of metal is-

A current of 2 ampere was passed through solutions of CuSO₄ and AgNO₃ in series. 0.635 g of copper was deposited. Then the weight of silver deposited will be-

A solution containing one mol per litre each of Cu(NO3)2, AgNO3, Hg2(NO3)2 and MgSO4 is being electrolysed by using inert electrodes.The standard electrode potentials (reduction potentials) are: Ag⁺/Ag = +0.80 V Hg₂²⁺/Hg = +0.79 V Cu²⁺/Cu = +0.34 V M

1 Faraday of electricity will liberate 1 mole of the metal from the molten of-

How long 2.5 amp of current is passed to supply 54000 C of charge-

How many coulombs of electricity are required for the oxidation of 1 mole of H₂O to O₂-

What weight of copper (At. mass 63.5) deposits when 2 Faraday of electricity is passed through cupric salt-

The number of faraday required to generate 1 mole of Mg from MgCl₂ is-

One faraday of electricity will liberate one mole of metal from a molten of-

The unit of electrochemical equivalent is-

The quantity of electricity required to liberate 0.01 g equivalent of an element at the electrode is-

Three Faradays of electricity are passed through molten AlCl₃, aqueous solution of CuSO₄ and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of-

A current of 2.6 ampere is passed through CuSO₄ solution for 6 minutes 20 seconds. The amount of Cu deposited is (At. wt. of Cu = 63.5, Faraday = 96500 C)-