A current of 3 ampere passed for 2 hours through a solution of CuSO4 3 gm of Cu2+ ions were discharged at cathode calculate the current efficiency.

 

Related Questions:

(1) The conductivity of 0.0011028 mol per liter acetic acid is 4.95 × 10–5 S per cm. Calculate its dissociation constant if ^°m for acetic acid is 390.5 S cm2 per mol–1 .

(2) In the electrolysis of aqueous solution of NaOH , 2.8 litre of oxygen at NTP liberated at anode. How much Hydrogen was liberated at cathode ?

(3) On passing 3 faradays of electricity through three electrolytic cells connect in series containing Ag+ , Ca+2 and Al+3 ions respectively , the molar ratio in which three metals ions are liberated at the electrode is

(4) How to determine anode and cathode of concentration cell instead of both electrodes are same ?

(5) Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]

(6) A silver electrode is immersed in saturated solution of Ag2SO4(aq). The potential difference between the silver and standard electrode is found to be 0.711 V. determined Ksp(Ag2SO4) given E0Ag+/Ag= 0.799 V

(7) How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hours. At a current of 8.46 ampere? What is the area of tray if the thickness of silver plating is 0.00254 cm. density of silver is 10.5 gm cc.

(8) Calculate the emf of the cell: Pt H2(1atm)ICH3COOH(0.1M) II NH4OH(0.01M)IH2(1 atm)Pt and Ka for CH3COOH= 1.8x10^-5 and Kb for NH4OH = 1.8x10^-5

Calculate the emf of the cell: Pt H2(1atm)ICH3COOH(0.1M) II NH4OH(0.01M)IH2(1 atm)Pt and Ka for CH3COOH= 1.8x10^-5 and Kb for NH4OH = 1.8x10^-5

 

(A) Reaction at anode:

(B) Reaction at Cathode:

From equation (1)

Alternative Method:

(A) Reaction at anode:

(B) Reaction at Cathode:

From equation (1)

Related Questions:

(1) The conductivity of 0.0011028 mol per liter acetic acid is 4.95 × 10–5 S per cm. Calculate its dissociation constant if ^°m for acetic acid is 390.5 S cm2 per mol–1 .

(2) In the electrolysis of aqueous solution of NaOH , 2.8 litre of oxygen at NTP liberated at anode. How much Hydrogen was liberated at cathode ?

(3) On passing 3 faradays of electricity through three electrolytic cells connect in series containing Ag+ , Ca+2 and Al+3 ions respectively , the molar ratio in which three metals ions are liberated at the electrode is

(4) How to determine anode and cathode of concentration cell instead of both electrodes are same ?

(5) Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]

(6) A silver electrode is immersed in saturated solution of Ag2SO4(aq). The potential difference between the silver and standard electrode is found to be 0.711 V. determined Ksp(Ag2SO4) given E0Ag+/Ag= 0.799 V

(7) How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hours. At a current of 8.46 ampere? What is the area of tray if the thickness of silver plating is 0.00254 cm. density of silver is 10.5 gm cc.

(8) A current of 3 ampere passed for 2 hours through a solution of CuSO4 3 gm of Cu2+ ions were discharged at cathode calculate the current efficiency.

How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8 hours. At a current of 8.46 ampere? What is the area of tray if the thickness of silver plating is 0.00254 cm. density of silver is 10.5 gm cc.

 

Related Questions:

(1) The conductivity of 0.0011028 mol per liter acetic acid is 4.95 × 10–5 S per cm. Calculate its dissociation constant if ^°m for acetic acid is 390.5 S cm2 per mol–1 .

(2) In the electrolysis of aqueous solution of NaOH , 2.8 litre of oxygen at NTP liberated at anode. How much Hydrogen was liberated at cathode ?

(3) On passing 3 faradays of electricity through three electrolytic cells connect in series containing Ag+ , Ca+2 and Al+3 ions respectively , the molar ratio in which three metals ions are liberated at the electrode is

(4) How to determine anode and cathode of concentration cell instead of both electrodes are same ?

(5) Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]

(6) A silver electrode is immersed in saturated solution of Ag2SO4(aq). The potential difference between the silver and standard electrode is found to be 0.711 V. determined Ksp(Ag2SO4) given E0Ag+/Ag= 0.799 V

A silver electrode is immersed in saturated solution of Ag2SO4(aq). The potential difference between the silver and standard electrode is found to be 0.711 V. determined Ksp(Ag2SO4) given E0Ag+/Ag= 0.799 V

Related Questions:

(1) The conductivity of 0.0011028 mol per liter acetic acid is 4.95 × 10–5 S per cm. Calculate its dissociation constant if ^°m for acetic acid is 390.5 S cm2 per mol–1 .

(2) In the electrolysis of aqueous solution of NaOH , 2.8 litre of oxygen at NTP liberated at anode. How much Hydrogen was liberated at cathode ?

(3) On passing 3 faradays of electricity through three electrolytic cells connect in series containing Ag+ , Ca+2 and Al+3 ions respectively , the molar ratio in which three metals ions are liberated at the electrode is

(4) How to determine anode and cathode of concentration cell instead of both electrodes are same ?

(5) Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]

How to calculate packing efficiency of cubic unit cell of diamond ?

Diamond structure is ZnS type structure  in which carbon atoms forms  a face centred cubic (FCC/CCP) lattice as well as  four out of eight (50%) or alternate tetrahedral voids are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four other. No discrete molecule can be discerned (identified) in diamond .the entire crystal is giant molecule a unit cell of which is shown as below.

Note : Only those atoms which form four covalent bond produce a repeated 3D structure using only covalent bonds.

Lattice of Diamond is ZnS type structure.

(1) C- form FCC/CCP (4-atoms)

(2) C- atoms present at the (50%) alternative tetrahedral voids (4-atoms)

(3) Total Number of one lattice unit is eight (8) hence molecular formula of diamond is (C8) (i.e. Z= 8)

(4) Number of C-C bond in lattice cell is = 4×4= 16

(5) Number of C-C bond per carbon atom is 16/8=2

(6) The distance between two Corbin atom is d_C-C = a√3/4 and the radius of carbon atom = dc-c/2 = r_c = a√3/2x4

(8) Packing efficiency (PE = π√3/10= 0.34 or 34%):

(9) Voids = 66 % 

Additional Information: 

Related Questions:

What are the normal spinel structures?

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How to calculate radius ratio of square Voids?

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How to calculate packing fraction or packing efficiency of two dimensional (2D) hexagonal packing solid atoms?

The coordination number of Al in the crystalline state of AlCl3 is .... (IIT-JEE 2009)

The number of hexagonal faces that are present in truncated octahedron is .... (IIT-JEE 2011)

What are structural information of Dimomd ?

In diamond, carbon atoms occupy FCC/CCP lattice point as well as alternate tetrahedral voids. If edge length of unit cell is 3.56pm, the radius of carbon atom is ?

The edge length of unit cubic cell of diamond is 356.7 pm and this cubic unit cell contains 8 carbon atoms calculate (A) Distance between two carbon atoms, assuming them to spheres in contact (B) Radius of carbon atoms (C) Fraction of the total volume that is occupied by the carbon atoms.