Calculate the cell potential E cell of the following cell at 298 K . Ag ( s ) I AgNO 3 0 . 01 M ∥ AgNO 3 1 . 0 M I Ag s

Given, Ag s I AgNO 3 0 . 01 M ∥ AgNO 3 1 . 0 M I Ag s When both the electrode are of same metal, then E cell ° = 0 We know that, E cell = E cell ° - 0 . 0591 n log Product Reactant E cell = 0 - 0 . 0591 1 log 0 . 01 1 = 0 - 0 . 0591 × - 2 log 10 = 0 . 0591 × 2 = 0 . 1182 V

How much amount of substance is deposited by passing one Faraday of electricity?

One Faraday = 96500 C of electricity deposits one gram equivalent 1 g Eq . of the substance because the equivalent weight of any element is the ratio of the atomic weight of the element to the number of electrons gained or lost by one atom/ion of the element. Therefore, we can conclude that Z = Eq . wt . of the substance 96500 . Here Z is the electrochemical equivalent and is given by W Q . Therefore, W Q = Eq . wt . of the substance 96500 , where W is the weight of the substance deposited in W gram on passing Q (Coulombs of electricity).

Explain Normal or Standard hydrogen electrode.

The potential of individual half-cells cannot be measured. So, we can measure only the difference between the two half-cell potentials that gives the emf of the cell. If we arbitrarily choose the potential of one electrode (half-cell) then that of the other can be determined with respect to this. According to the convention, a half-cell called standard hydrogen electrode as shown below represented by Pt s | H 2 g | H + aq , is assigned a zero potential at all temperatures corresponding to the reaction. H + aq + e - → 1 2 H 2 g . The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity. This implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ions in the solution is one molar.

Give two differences between E.M.F and potential difference.

Electro-Motive Force (EMF) Potential Difference In an open circuit, it is the difference in potential between the two electrodes of a cell or battery. It is the difference of potential between any two points in a closed circuit Electro-Motive Force is always greater than the potential difference between any two points of a conductor. The Electro-Motive Force is always less than the potential difference.

What is corrosion? Explain two factors which affect corrosion.

Corrosion is the damage caused to the metal by the chemical reaction of air, water and acids with the surface of the metal. Generally reactive metals corrode easily and non-reactive metals have good resistance to corrosion. For example, Copper reacts with moist carbon dioxide in the air and gets a green coat of copper carbonate. The two factors which affect corrosion are: More the reactivity of metal, the more will be the possibility of the metal getting corroded. Presence of electrolytes in water increases the rate of corrosion.

The molar conductivity at infinite dilution for Sodium Acetate CH 3 COONa , Hydrochloric Acid HCl and Sodium Chloride NaCl are 91 . 0 , 426 . 2 and 126 . 5 Scm 2 mol - 1 respectively. Calculate the molar conductivity at infinite dilution of Acetic Acid CH 3 COOH ?

Given, Λ m o CH 3 COONa = 91 Scm 2 mol - 1 , Λ m o HCl = 425 . 9 Scm 2 mol - 1 , Λ m o NaCl = 126 . 4 Scm 2 mol - 1 Applying Kohlrausch's law, the molar conductivity of acetic acid is calculated as: Λ m o CH 3 COOH = Λ m o CH 3 COO - + Λ m o H + = Λ m o CH 3 COO - + Λ m o Na + + Λ m o H + + Λ m o Cl - - Λ m o Na + + Λ m o Cl - = Λ m o CH 3 COONa + Λ m o HCl - Λ m o NaCl = 91 + 425 . 9 - 126 . 4 = 390 . 5 Scm 2 mol - 1

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Electrochemistry>Punjab Board-2019>Q 6

MEDIUM

12th Punjab Board

IMPORTANT

Earn 100

Calculate the molar conductance of a solution of ${MgCl}_{2}$ at infinite diltution given that the molar ionic conductance of $λ^{o} ({Mg}^{+ 2}) = 126.1 {Scm}^{2} {mol}^{- 1}$ and $λ^{o} ({Cl}^{-}) = 56.3 {Scm}^{2} {mol}^{- 1}$ .

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