In the button cells widely used in watches and other devices the following reaction takes place: Zn s + Ag 2 O s + H 2 O l → Zn 2 + aq + 2 Ag s + 2 OH - aq Determine ∆ r G and E ° cell for the reaction. E ° Zn 2 + / zn = — . 76 V , E ° Ag + / Ag = + . 80 V

The following reactions take place in the button cells: Anode reaction is: Z n s → Z n a q 2 + + 2 e - ; E 0 = - 0.76 V Cathode reaction is: A g 2 O s + H 2 O l + 2 e - → 2 A g s + 2 O H - a q ; E 0 = 0. 80 V Overall reaction: Z n s + A g 2 O s + H 2 O l → Z n 2 + a q + 2 A g s + 2 O H - a q As we know, E c e l l ∘ = E c a t h o d e ∘ - E a n o d e ∘ E ∘ cell = 0 . 80 + 0 . 76 = 1 . 56 V Now, the relation between standard Gibb's free energy and standard cell potential is, ∆ r G ∘ = - n F E ∘ = - 2 × 96500 × 1. 56 = - 301080 J mol - 1 = - 301 .08 k J mol - 1

Rusting of iron is said to be an electrochemical phenomenon. Give reason for this.

In corrosion, a metal is oxidised by the loss of electrons to oxygen and the formation of oxides. Corrosion of iron (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon because at a particular spot of an object made of iron, the oxidation takes place and that spot behaves as an anode, and we can write the reaction(it has given below) and it involves the oxidation of metal which act as an anode and the reduction of atmospheric gases in the presence of water and is called a cathode. Oxidation at the anode: 2 Fe ( s ) → 2 Fe 2 + + 4 e – . Reduction at the cathode : O 2 ( g ) + 4 H + ( aq ) + 4 e – → 2 H 2 O ( l ) . The overall reaction being: 2 Fe ( s ) + O 2 ( g ) + 4 H + ( aq ) → 2 Fe 2 + ( aq ) + 2 H 2 O ( l ) . Thus iron loses electrons and get converted to rust and lose its shine because of electrochemical reaction.

What is sacrificial protection?

Sacrificial protection: It is defined as corrosion protection method in which a more electrochemically active metal is electrically attached to a less active metal and this reverts the protected metal back to its original form, and thereby prevents it from corroding.

What are fuel cell ? Write reaction taking place in H 2 - O 2 fuel cell.

Galvanic cells which convert the energy of combustion of fuels like hydrogen, methanol, methane, etc. directly into electrical energy are called fuel cells. The electrode reaction takes place in H 2 - O 2 fuel cells are following, Cathode: O 2 g + 2 H 2 O l + 4 e - → 4 OH - aq Anode: 2 H 2 g + 4 OH - aq → 4 H 2 O l + 4 e - Overall reaction: 2 H 2 g + O 2 g → 2 H 2 O l

Mercury cell gives a constant voltage of 1 . 35 V during its lifetime. Why?

Mercury cell gives a constant voltage of 1 . 35 V during its lifetime and remains constant also because the overall reaction does not involve any ion in a solution whose concentration can change during its lifetime. Zn ( Hg ) + HgO ( s ) → ZnO ( s ) + Hg ( l )

Explain giving reasons : The E ° M 2 + / M value for copper is positive (0.34 V).

E ° M 2 + / M value of a metal depends on the energy changes with the sublimation, ionization and hydration etc., Sublimation: It is the energy required for converting one mole of an atom from the solid-state to the gaseous state. Ionization: It is the energy required to take out electrons from one mole of atoms in the gaseous state. Hydration: It is the energy released when one mole of ions is hydrated. Here, copper has a high energy of atomization and low hydration energy. Therefore, the E ° M 2 + / M value for copper is positive.

Λ ° m for NaCl , HCl and NaAc are 126 . 4 , 425 . 9 and 91 . 05 cm 2 mol - 1 respectively. Calculate Λ ° m for HAc.

Given, Limiting molar conductivity of NaCl , ( Λ ° m ) = 126 . 4 S cm 2 mol - 1 Limiting molar conductivity of HCl , ( Λ ° m ) = 425 . 9 S cm 2 mol - 1 Limiting molar conductivity of NaAc , ( Λ ° m ) = 91 . 0 S cm 2 mol - 1 By applying Kohlrausch's law, ( Λ ° m ) HAc = λ ° H + + λ ° Ac - . . . ( 1 ) ( Λ ° m ) NaCl = λ ° Na + + λ ° Cl - . . . ( 2 ) ( Λ ° m ) HCl = λ ° H + + λ ° Cl - . . . ( 3 ) ( Λ ° m ) NaAc = λ ° Na + + λ ° Ac - . . . . ( 4 ) Adding ( 3 ) and ( 4 ) and subtracting ( 2 ) we get, ( Λ ° m ) HAc = ( Λ ° m ) HCl + ( Λ ° m ) NaAc - ( Λ ° m ) NaCl = ( 425 . 9 + 91 . 05 ) - 126 . 4 = 390 . 55 S cm 2 mol - 1 Hence, the molar conductivity ( Λ ° m ) for HAc is 390 . 55 S cm 2 mol - 1 .

State Faraday’s first law of electrolysis.

According to Faraday’s first law of electrolysis, The amount (mass) of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of charge passing through the electrolyte (solution or melt). W ∝ Q W = Z × i × t Where, W = Mass of ions liberated in gm Q = Quantity of electricity passed in Coulombs i = Current in amperes t = Time in seconds And Z = Constant which is known as equivalent of the ion deposited.

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While giving labelled diagram of dry cell write reactions taking place at cathode and anode.

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