When an aqueous solution containing gold (III) salt is electrolyzed, metallic gold is deposited at the cathode, and oxygen gas is generated ?

When an aqueous solution containing gold (III) salt is electrolyzed, metallic

gold is deposited at the cathode and oxygen gas is generated at the anode. If  

9.26 g of Au is deposited at the cathode, 

calculate the volume (in liters) of O2 generated at 23°C and 747 mmHg.

1 Answer

Relevance
  • Bobby
    Lv 7
    1 month ago
    Favourite answer

    at the cathode we have 

    Au3+ + 3e- = Au 

    we have 9.26 g or 9.26 g / 197 g / mol = 4.70* 10 ^-2 moles of Au plated

    moles of electrons plating this gold according to our equation 

    = 4.70* 10^-2 *3 = 0.141 moles of electrons 

    H2O = O2+ 4H+ + 4e-

    moles of O2 released by 0.141 moles of electrons 

    =  0.141 moles /4 = 0.0352 moles of O2 

    volume of these moles 

    = n *R *T / P = 0.0352 moles * 62.36367 L mmHg K−1 mol−1 * 296 K / 747  mmHg.

    = 0.867 L of O2 

    Source(s): of Au plated
Still have questions? Get answers by asking now.