100 mL of 0.10 M CuSO4 is electrolysed between platinum electrodes with a steady current of 0.50 A. Take F = 96500 C mol

Question

100 mL of 0.10 M CuSO4 is electrolysed between platinum electrodes with a steady current of 0.50 A. Take F = 96500 C mol^-1 and neglect volume changes. Which statement is correct?

Accepted Answer

After 3860 s the cathode product changes from Cu to H2 and the solution is then 0.10 M in H2SO4

Answer

After 1930 s the cathode product changes from Cu to H2 and the solution is then 0.05 M in H2SO4

Answer

After 3860 s oxygen starts to evolve at the cathode

Answer

Copper deposition continues indefinitely because Cu^2+ is regenerated at the anode

$100 mL$ of $0.10 M$ $CuSO X_{4}$ is electrolysed between platinum electrodes with a steady current of $0.50 A$ . Take $F = 96500 C mo l^{- 1}$ and neglect volume changes. Which statement is correct?

Answer: A

More questions on Electrolysis and products at electrodes

100 mL of 0.10 M CuSOX4​ is electrolysed between platinum electrodes with a steady current of 0.50 A. Take F=96500 Cmol−1 and neglect volume changes. Which statement is correct?

Answer: A

More questions on Electrolysis and products at electrodes

$100 mL$ of $0.10 M$ $CuSO X_{4}$ is electrolysed between platinum electrodes with a steady current of $0.50 A$ . Take $F = 96500 C mo l^{- 1}$ and neglect volume changes. Which statement is correct?