electrochemistry ( electrolysis of metals)

1. Define Faraday's first law of electrolysis 2. define Faraday's second law of electrolysis

Related Questions in Chemistry

• Q :Determining of normality of sodium

Can someone please help me in getting through this problem. The normality of a solution of sodium hydroxide 100 ml of which includes 4 grams of NaOH is: (a) 0.1 (b) 40 (c) 1.0 (d) 0.4

• Q :Meaning of molality of a solution The

The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m

• Q :Diffusion Molecular View When the

When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.To see how the experimental coefficients can be treat

• Q :Molarity of acid solution If 20ml of

If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M  (c) 0.3M (d) 0.4M Choose the right answer fron above.

• Q :Define the term oxidizing agent Briefly

Briefly define the term oxidizing agent?

• Q :How to calculate solutions ionic

Transference numbers and molar conductors can be used to calculate ionic mobilities. This tables under is giving the transference numbers for positive ions at 25 degree C and the values obtained by extrapolation to infinite dilution:

Q :Explain the molecular mass with respect

During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is

• Q :Question on Mole fraction Mole fraction

Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of  moles of solute/ Mass of solvent in kg (d) no. of moles of any

• Q :Question based on lowering of vapour

Choose the right answer from following. The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be:  (a) 18.0 (b) 342 (c) 60 (d) 180

• Q :Influence of temperature Can someone

Can someone please help me in getting through this problem. With increase of temperature, which of the following changes: (i) Molality (ii) Weight fraction of solute (iii) Fraction of solute present in water (iv) Mole fraction.