Electrochemistry MCQs for NEET — Chemistry Questions with Answers

For the half-cell reaction involving hydrogen: \[ Pt | H_2 (1 ext{ bar}) | H^+ (a = 1) \] The oxidation potential can be calculated using the Nernst equation: \[ E = E^ ext{°} + rac{0.0591}{n} ext{log} rac{[H^+]}{P_{H_2}} \] Given that \(PH = 11\), which means the concentration of \(H^+\) is \(10^{-11}\). Plugging in the values, we get the oxidation potential as 0.177 V.

In an electrochemical cell, when the concentration of the solution in the cathode increases, the reaction quotient Q decreases because the concentration of the reactants increases. According to the Nernst equation, $E_{cell} = E^{ ext{cell}} - rac{RT}{nF} ext{ln} Q$, a decrease in Q results in a decrease in the cell potential, E. Therefore, the cell potential decreases.

A metal can displace another metal from its salt solution if it is a stronger reducing agent. Since an aqueous solution of a salt of metal B can be stored in a vessel of metal A, metal A is a stronger reducing agent than metal B ($A > B$). Similarly, since an aqueous solution of a salt of metal C can be stored in a vessel of metal B, metal B is a stronger reducing agent than metal C ($B > C$). Therefore, the order of their strength as reducing agents is $C > B > A$.

During the electrolysis of molten CaCl_2, calcium ions are reduced at the cathode. According to Faraday's laws of electrolysis, 2 Faradays of charge are required to deposit 1 mole of Ca. Therefore, 5 Faradays would theoretically deposit 2.5 moles of Ca. However, due to practical losses and inefficiencies, the actual amount obtained is less than 2.5 moles.

The number of moles of metal deposited during electrolysis is inversely proportional to the equivalent weight (molar mass divided by the valency). For Cu (Cu^2+), Ag (Ag^+), and Ni (Ni^2+), the valencies are 2, 1, and 2 respectively. Thus, the proportion of moles of metals deposited is inversely proportional to these valencies, giving the ratio 1:2:1.