Asked by sarah simpson
Electrochemistry
In the electrolysis of aqueous copper(II) bromine CuBr2(aq, 1.00 g of copper is deposited at the cathode. How many grams of bromine are produced at the anode?
In the electrolysis of aqueous copper(II) bromine CuBr2(aq, 1.00 g of copper is deposited at the cathode. How many grams of bromine are produced at the anode?
Answers
Answered by
DrBob222
96,485 C will deposit 63.54/2g Cu: therefore, how many C will deposit 1 g Cu? That's
96,485 x 2/63.54 = ?
Then 96,485 will deposit 160/2 g Br2. How man g will ? from above deposit.
That's (160/2) x ? from above = ?g Br2.
You should confirm all of the calculations AND confirm the atomic/molar masses of Cu and Br2.
96,485 x 2/63.54 = ?
Then 96,485 will deposit 160/2 g Br2. How man g will ? from above deposit.
That's (160/2) x ? from above = ?g Br2.
You should confirm all of the calculations AND confirm the atomic/molar masses of Cu and Br2.
Answered by
Chemman7
Since 1 g of copper is deposited, that is 1/63.55 = 0.016 mol of Cu approximately.
Since the equation is CuBr2 -> Cu2+ + 2 Br- the mol ratio is 2 to 1.
2:1
Br:Cu
x:0.16
x=0.0315 mols of Br.
since m = nM,
m= 0.0315 x 79.9 (Molar mass of Br)
m=2.5145 g of Br
Since the equation is CuBr2 -> Cu2+ + 2 Br- the mol ratio is 2 to 1.
2:1
Br:Cu
x:0.16
x=0.0315 mols of Br.
since m = nM,
m= 0.0315 x 79.9 (Molar mass of Br)
m=2.5145 g of Br
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