Out of zinc and tin, whose coating is better to protect iron objects?

Coating of zinc metal on the surface of iron is called galvanization. When iron object is coated with zinc, a thin layer of zinc solidifies on the surface of the iron. In galvanized iron the zinc acts as anode in the presence of air and water. This is because zinc is more reactive than iron. Whereas, tin has lower oxidation potential than that of zinc, and it has less susceptibility to abrasion and mechanical damage. Hence, zinc coating is better to protect iron objects.

Give two advantages of fuel cells.

Fuel cells are defined as the galvanic cells in which the energy of combustion of fuels is directly converted into electrical energy. One of the reactants is fuel such as hydrogen or methanol. Advantages of fuel cells are: (i) They are non-polluting because the only reaction product is water. (ii) The reactants are continuously supplied to the electrodes from the reservoir. Unlike conventional cells, the fuel cells do not have to be discharged when the chemicals are consumed. They also have high efficiency.

Calculate ∆ G ° for the reaction Zn s + Cu 2 + aq → Zn 2 + aq + Cu s Given: E ° for Zn 2 + / Zn = - 0 . 76 V E ° for Cu 2 + / Cu = + 0 . 34 V R = 8 . 314 JK - 1 mol - 1 F = 96500 C mol - 1

Given reaction is Zn s + Cu 2 + aq → Zn 2 + aq + Cu s We know that, E cell ° = E cathode ° - E anode ° E cell ° = 0 . 34 - - 0 . 76 E cell ° = 1 . 10 V The standard Gibbs free energy change is the energy change that occurs in going from the reactants to the products. It can be calculated by using the formula, ∆ G ° = - nFE cell ° ∆ G ° = - 2 × 1 . 10 × 96500 = - 212 . 3 kJ mol - 1

In the plot of molar conductivity ( ∧ m ) vs square root of concentration ( c 1 / 2 ) , following curves are obtained for two electrolytes A and B : Answer the following : What happens on extrapolation of ∧ m to concentration approaching zero for electrolytes A and B ?

On extrapolation, for electrolyte, A limiting value of conductance is obtained that is conductance at zero concentration or infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at infinite dilution (zero concentration).

In the plot of molar conductivity ( ∧ m ) vs square root of concentration ( c 1 / 2 ) , following curves are obtained for two electrolytes A and B : Answer the following : Predict the nature of electrolytes A and B.

From the given graph we can say that: Electrolyte A : Strong electrolyte For strong electrolytes, the molar conductivity increases slowly with the dilution. The plot between the molar conductivity and c 1 / 2 is a straight line Electrolyte B : Weak electrolyte For weak electrolytes, the graph plotted between molar conductivity and c 1 / 2 (where c is the concentration) is not a straight line. Weak electrolytes have lower molar conductivities and lower degree of dissociation at higher concentrations which increases steeply at lower concentrations.

A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO 4 and ZnSO 4 until 2 . 8 g of Fe deposited at the cathode of cell X. How long did the current flow ? Calculate the mass of Zn deposited at the cathode of cell Y. ( Molar mass : Fe = 56 g mol - 1 Zn = 65 . 3 g mol - 1 , 1 F = 96500 C mol - 1 )

Given that current, I = 2 A Reaction: Fe 2 + + 2 e - → Fe Thus, 56 g Fe requires 2 × 96500 C charge. Hence, 2 . 8 g Fe will require 2 × 96500 56 × 2 . 8 = 9650 C charge. As we know, Q = It ; t = Q I ⇒ t = 9650 2 = 4825 sec Thus, the current flowed till 4825 seconds . As, Zn 2 + + 2 e - → Zn 2 × 96500 C deposits 65 . 3 g Zn . 9650 C deposits : 65 . 3 2 × 96500 × 9650 = 3 . 265 g Zn .

E o cell for the given redox reaction is 2 . 71 V . Mg ( s ) + Cu 2 + ( 0 . 01 M ) → Mg 2 + ( 0 . 001 M ) + Cu ( s ) Calculate E cell for the reaction. Write the direction of flow of current when an external opposite potential applied is greater than 2 . 71 V .

According to Nernst equation: E cell = E o cell - 2 . 303 RT nF log C 2 C 1 On substituting the given values: E cell = 2 . 71 - 2 . 303 × 8 . 314 × 298 2 × 96500 log 0 . 001 0 . 01 = 2 . 74 V When an external opposite potential is applied more than 2 . 71 V the direction of flow of current would be reversed.

E o cell for the given redox reaction is 2 . 71 V . Mg ( s ) + Cu 2 + ( 0 . 01 M ) → Mg 2 + ( 0 . 001 M ) + Cu ( s ) Calculate E cell for the reaction. Write the direction of flow of current when an external opposite potential applied is less than 2 . 71 V .

According to Nernst equation: E cell = E o cell - 2 . 303 RT nF log C 2 C 1 On substituting the given values: E cell = 2 . 71 - 2 . 303 × 8 . 314 × 298 2 × 96500 log 0 . 001 0 . 01 = 2 . 74 V When an external opposite potential is applied less than 2 . 71 V the direction of flow of current would remain same.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Electrochemistry>CBSE-2020>Q 18

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12th CBSE

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Give two points of differences between electrochemical and electrolytic cells.

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