Chemistry NCERT Exemplar Solutions Class 12th Chapter Three

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Ans: Electrolysis of brine solution is as follows

NaCl (aq)→Na⁺ (aq)+Cl⁻ (aq)

At the cathode, the reaction goes this way,

H₂O (l) + e⁻→12H₂ (g)+OH⁻ (aq)

At the anode, the reaction goes this way,

Cl⁻ (aq)→12Cl₂ (g) + e⁻

The pH of the solution will increase as sodium hydroxide is being formed.

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Ans: When the opposing potential becomes equal to the electrical potential there is no current flowing in the cell and the cell reaction stops and there is no chemical reaction in the cell

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Ans: Alternating current is used to stop electrolysis so that the concentration of the ions in the solution remains constant.

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Ans: The cell represented above shows an electrochemical cell in which two different electrodes are present and the cell at the bottom represents the electrolytic cell.

In this zinc is losing electrons moving towards electrode A and copper is accepting an electron from electrode B. Therefore, the polarity of electrode A is positive and electrode B is negative.

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Ans: The potential difference between the metal and its solution is termed electrode potential.

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Ans: In the electrolysis process of sodium chloride, oxidation of water at anode requires more potential. So Cl− is oxidized at anode instead of water.

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Ans:  Galvanic cell consists of two electrodes in which anode and cathode are present as electrodes. The anode is present on the left side at which oxidation occurs. The cathode is present on the right side at which reduction occurs and in the middle, there is a salt bridge which is depicted by parallel lines. Therefore, the galvanic cell is Cu|Cu²⁺ |Ag⁺ |Ag.

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Ans: According to Faraday's second law of electrolysis, the amount of different substances liberated keeping the electricity flow same through the electrolytic solution is directly proportional to the equivalent weight.

$\frac{{\mathrm{\omega }}_1}{{\mathrm{\omega }}_2}$ = $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}$

E₁and E₂ have different values. Therefore, the mass of copper and silver deposited will be different

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Ans: At the stage of equilibrium E? _Cell =0. So according to the relation,

ΔrG =−nFEcell

ΔrG =−n*F*0

ΔrG =0

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Ans: E? _Cell can never be zero. For a feasible reaction E? _Cell should be positive or ΔrG$°$should be negative and at the stage of equilibrium, both of these parameters are zero.