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If \( E^\circ_{Fe^{2+}/Fe} = -0.441 \, \text{V} \) and \( E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \, \text{V} \),
the standard emf of the cell reaction \( Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \) is:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction, \( Fe^{3+} \) is reduced to \( Fe^{2+} \) (reduction at the cathode), and \( Fe \) is oxidized to \( Fe^{2+} \) (oxidation at the anode). So: \[ E^\circ_{\text{cell}} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Fe^{2+}/Fe} \] \[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 0.771 + 0.441 = 1.212 \, \text{V} \] Hence, the standard emf of the cell reaction is \( 1.212 \, \text{V} \).

  • AP EAPCET - 2025
  • AP EAPCET
  • Chemistry
  • Electrolysis

Consider the following
Statement-I: Kolbe's electrolysis of sodium propionate gives n-hexane as product. 
Statement-II: In Kolbe's process, CO$_2$ is liberated at anode and H$_2$ is liberated at cathode.

  • AP EAPCET - 2025
  • AP EAPCET
  • Chemistry
  • Electrolysis
96.5 amperes current is passed through the molten AlCl$_3$ for 100 seconds. The mass of aluminum deposited at the cathode is (Atomic weight of Al = 27 u)
  • AP EAPCET - 2022
  • AP EAPCET
  • Chemistry
  • Electrolysis