Question

A copper-nickel (35 wt % - 65 wt %) alloy is corroded in an oxygen-concentration cell using boiling water. What volume of oxygen gas (at 1 atm) must be consumed at the cathode to corrode 54 g of the alloy? (Assume only divalent ions are produced.) Express your answer to three significant figures.

Answer 1

The corrosion reaction

and the gaseous reduction reaction is the production (and consumption) of electrons:

Cu ---> Cu²⁺ + 2e-                   

and

O₂ + 2H₂O-----> 4e^- + 4OH^-.

One mole of Cu produces 2 moles of electrons, but only 1 mole of
Cu²⁺
O2 gas is needed to consume 2 moles of electrons. Using data from
, we can write

#moles of O2 gas = 1/2 mole Cu

mass of Cu 0.35*54 =18.9 g

molar mass of Cu is 63.55 g/mol

moles of Cu = mass of Cu/Molar mass of Cu

=18.9/63.55

=0.297 moles Cu

Thus the No. of Cu present is 0.297 mole Cu

so , 0.297/2 moles of O₂ will be required for corrosion

0.297/2 = 0.1487 moles of O₂

volume of Oxygen gas can be found by ideal gas law

V=nRT/P

=0.1487*8.314*(373)/101325

=0.00455m³

this is how we can solve for alloy

similarly for Ni

Ni in layer = 0.65*54=35.1 g

moles of Ni = mass /molar mass =35.1/58.7=0.5979 moles of Ni

each mole of Ni will requires 0.5 moles of O₂

Oxygen required = 0.2989 moles

Volume = 0.2989*8.314*373/101325

=0.00914 m³

total volume of oxygen needed = 0.00455+0.00914=0.01369 m³