Answer:
3.16g
Explanation:
Data obtained from the question include:
I = 4A
t = 40mins = 40 x 60 = 2400secs
Q =?
Q = It
Q = 4 x 2400
Q = 9600C
Now let us generate a balanced half equation to determine the number faraday needed to deposit metallic Cu. This is illustrated:
Cu^2+ + 2e —> Cu
From the equation,
2 faraday is needed to deposit metallic cu
1 faraday = 96500C
Therefore 2 faraday = 2 x 96500 = 193000C
Molar Mass of Cu = 63.5g/mol
193000C deposited 63.5g of Cu
Therefore, 9600C will deposit = (9600 x 63.5)/193000 = 3.16g of Cu
