Electrolysis of dil H2SO4 liberates gases at anode and cathode:
Correct answer: C. O2 & H2 respectively
- A. O2 & SO2 respectively
- B. SO2 & O2 respectively
- C. O2 & H2 respectively
- D. H2 & O2 respectively
Explanation
In the electrolysis of dilute sulfuric acid (H2SO4), water is decomposed into oxygen and hydrogen gases. At the cathode, hydrogen ions (H+) gain electrons (reduction) to form hydrogen gas (H2):2H+ + 2e- → H2At the anode, water molecules are oxidized to produce oxygen gas (O2):2H2O → O2 + 4H+ + 4e-Thus, oxygen is liberated at the anode and hydrogen is liberated at the cathode. Options A and B are incorrect as they suggest SO2 is produced, which does not occur in this reaction. Option D incorrectly reverses the roles of the gases at the electrodes.
Last updated
About Electrochemistry
Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.
Practise Electrochemistry
696 free Electrochemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
10-2 mole of Fe3O4 is treated with excess KI solution in presence of dilute H2SO4, the products are Fe2+ and I2(g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2(g)?
+3 and +5 oxidation states are often shown by:
3 faraday of electricity is passed through molten Al2O3 , aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of:
5.3 gm of M2CO3 is dissolved in 150 ml of 1N HCl. The unused acid required 100 ml of 0.5N NaOH. Hence equivalent weight of M is:
8H++ MnO4- → Mn2+ + 4H2O , which one is correct about given equation?