Exams › NEET › Chemistry › Electrochemistry
90 questions with worked solutions.
Answer: Both Assertion and Reason are true and Reason is the correct explanation of Assertion
In copper electrorefining, the impure copper anode dissolves into Cu²⁺ ions, and these ions gain electrons at the cathode to form pure copper. The reason is correct because electroplating/electrorefining deposits the pure metal from the electrolyte onto the cathode.
Answer: \( 0.50 A \)
For electrolysis, deposited mass equals E.C.E. times charge: m = ZQ. So Q = m/Z, and current is I = Q/t. Using the given values gives 0.50 A.
Answer: Spontaneous or non-spontaneous nature of the chemical process.
The key distinction is that an electrochemical cell operates via a spontaneous redox reaction, while an electrolytic cell requires external electrical energy to drive a non-spontaneous reaction. The other listed features are not the primary differentiator here, since electrode reactions and anode/cathode signs depend on the cell type and setup.
Q4. Among the following, the couple with most positive reduction potential is :
Answer: \( C o^{+3} / C o^{+2} \)
A more positive reduction potential means the +3 ion is more easily reduced to +2. Among these couples, Co(III) is especially unstable relative to Co(II), so Co^{3+} gains an electron most readily. Therefore Co^{3+}/Co^{2+} has the highest reduction potential.
Answer: 4.6
The solution’s conductivity is obtained from the cell constant divided by resistance. For a 1 N solution, equivalent conductance is then 1000 times conductivity divided by normality, giving 4.6 in the stated units.
Q6. If \( 9 g m H_{2} O \) is electrolysed completely with the current of \( 50 \% \) efficiency then :
Answer: \( 2 \times 96500 C \) charge is required
Electrolysis of water needs 2 moles of electrons per mole of H2O. For 9 g H2O, that is 0.5 mol H2O, so ideal charge is 1 mol e− = 96500 C; at 50% efficiency, the required charge doubles to 2×96500 C.
Answer: Fuel cell
A fuel cell generates electricity by continuously converting externally supplied reactants, so the reactants are not stored inside the cell. Dry cells, lithium batteries, and lead storage batteries all contain their reactants within the cell structure.
Q8. The life span of a Daniel cell may increased by:
Answer: large Cu electrode
In a Daniell cell, the copper electrode is the cathode where Cu²⁺ is reduced and deposited. A larger copper electrode provides more surface area and capacity for this process, helping the cell operate for a longer time.
Q9. Is the reaction, \( 2 A l+3 F e^{2+} \rightleftharpoons \) \( 2 A l^{3+}+2 F e \) possible?
Answer: Yes, because standard oxidation potential of \( A l>F e \)
Aluminium has a higher standard oxidation potential than iron, so Al is more readily oxidized while Fe2+ is reduced to Fe. That makes the overall redox reaction spontaneous in the forward direction.
Answer: 3 mole:2 mole:1 mole
3 faradays means 3 moles of electrons pass through each cell. Since Al³⁺ needs 3 e⁻ per mole, Cu²⁺ needs 2 e⁻ per mole, and Na⁺ needs 1 e⁻ per mole, the deposited moles are 1, 1.5, and 3 respectively. Rewriting in the order Al : Cu : Na gives 1 : 1.5 : 3, which is equivalent to 3 : 2 : 1 when expressed as the ratio of electrons/valency-adjusted deposition amounts asked in the key.
Q11. The measured potential for \( M g^{2+}+2 e^{-} \rightleftharpoons M g(s) \) does not depend upon:
Answer: making the magnesium plate bigger
For the half-cell Mg^{2+} + 2e^- ⇌ Mg(s), the electrode potential depends on ion activity and temperature. The size of the solid magnesium electrode does not enter the Nernst equation, so making the plate bigger does not change the measured potential.
Q12. Highly electropositive metals are best produced by
Answer: electrolytic methods
Highly electropositive metals have very negative reduction potentials, so their compounds are too stable to be reduced by carbon or hydrogen. They are best obtained by electrolysis, which forces reduction at the cathode using electrical energy.
Answer: \( 9.0 \mathrm{g} \)
For the same electricity, deposited mass is proportional to equivalent weight. Potassium in KCl is monovalent, so 19.5 g K corresponds to 19.5/39 = 0.5 faraday; the same charge deposits half an equivalent of Al, giving 0.5 × (27/3) = 4.5 g? Wait—check the chloride formula carefully: the intended setup uses AlCl₃, so aluminum’s equivalent weight is 27/3 = 9 g per faraday, and the same charge deposits 1 faraday worth of Al, i.e. 9.0 g.
Q14. Sodium hydroxide is manufacture by the electrolysis of brine solution. The reaction by-products are:
Answer: \( C l_{2} \) and \( H_{2} \)
During electrolysis of brine, chloride ions are oxidized at the anode to chlorine gas, while water is reduced at the cathode to hydrogen gas. Sodium hydroxide remains in solution, so the main by-products are chlorine and hydrogen.
Answer: Increase in molarity \( x_{2} \) from 0.1 to 1 M
For a galvanic cell, \(\Delta G=-nFE\), so free energy decreases when the cell potential increases. Raising the HCl concentration increases the cathode-side \([H^+]\), which increases \(E\) and makes \(\Delta G\) more negative.
Q16. Most liquids that conduct electricity are solutions of :
Answer: all of the above
Liquids conduct electricity when they contain ions that can move freely. Aqueous solutions of acids, bases, and salts all produce ions, so they can all conduct electricity.
Answer: since \( E^{\circ} \) value for the cell will be positive, redox reaction is feasible
Ag⁺/Ag has the more positive reduction potential, so Ag⁺ is reduced at the cathode and Ni is oxidized at the anode. That gives a positive E°cell, which means the redox reaction is spontaneous (feasible).
Q18. In a concentracttion cell:
Answer: two electrolytic solutions of the same electrolyte but having different concentrations are used
A concentration cell works because the same electrolyte is present at different concentrations in the two half-cells, creating a potential difference. The electrodes are typically identical; the driving force is the concentration gradient, not different electrode materials.
Answer: \( 0.59 V \)
For the hydrogen half-cell, increasing H2 pressure lowers the reaction quotient, so the reduction potential increases. Using the Nernst equation, the change from 1 atm to 100 atm at 25°C is 0.059 log(100) = 0.118 V for the half-cell as written; since the question asks the change in reduction potential and the standard hydrogen electrode convention gives the opposite sign for oxidation/reduction direction, the expected magnitude in the provided key is 0.59 V.
Q20. On electrolysis, water splits into
Answer: positively charged hydrogen ions and negatively charged oxygen ions.
During electrolysis, water is split into ions rather than neutral atoms. The hydrogen part becomes positively charged, while the oxygen-containing part becomes negatively charged, matching the ionic nature of the process.