Question on Red-Ox and equivalence

[zut212]

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A solution of CuSO4 was electrolyzed using platinum electrodes. A volume of 6L of O2 gas measured at STP was liberated at positive electrode. What mass of copper was deposited at the negative electrode.

ANS: 34 g Cu deposited.

1 mol O2 = 4 equivalents
6 * 4
----- equivalents Cu X 31.8g Cu/equivalent = 34 g Cu deposited.
22.4L


Here is my question: How do they figure that 1 mol O2 = 4 equivalents?

 

[vsl5]

Copper is reduced (from +2 to 0) and oxygen is oxidized (-2 to 0). So in terms of electrons, it is balanced. But since CuSO4 has 4 oxygens and O2 has 2 oxygens, you need to balance oxygen on the product side (ie: you need twice as many moles of O2). So 6L/22.4L/mol =0.268 mol of O2, meaning you need 0.268 x 2 mole of CuSO4 (so 0.536 mol of Cu is deposited). 0.536 * 63.5g/mol Cu = 34g. I am not sure how you got 31.8g/equivalent

 

[zut212]

31.8 is the molecular weight of Cu DIVIDED by 2.

 

[Rabolisk]

The relevant equations are Cu2+ + 2e- -> Cu and 2H2O -> O2 + 4H+ + 4e-.

2 equivalent of copper (2 moles per mole of electron) and 4 equivalent of oxygen (4 moles per mole of electron). Twice as many moles of oxygen are formed as moles of copper are formed. Do the math, and you get the answer.

The key is to figure out the reactions that happen. The positive electrode in an electrolytic cell is the anode. Even if you did not know this, the question tells you that copper and O2 are formed. Copper can only be formed as a result of the reduction of copper cations, thus oxygen must have been oxidized. A CuSO4 solution has Cu2+, SO42-, and H2O. The sulfate does not react, because sulfur cation or sulfur atom would have to be formed.