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Sagot :
32.88grams of (Ag) silver would plate onto the pure electrode if a current of 6. 8A flowed through the cell for 72 minutes.
Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction:
Ag+(aq)+e→Ag(s).
By Using the equation
Q= I×t, where Q = charge
I = current = 6.8 A
t= time in seconds= 72×60=4320s
So,
Q=6.8×4320
Q = 29,376C
We see that 1 mole of electrons produces 1 mole of Ag from the half reaction.
1mole of electrons= 96,500C
xmoles of electrons= 29,376C
Cross multiplying;
29,376= 96,500 × x
Now , dividing both sides of the by 96,500
29,376/96,500=x
0.3044 moles of electrons will produce
0.3044×108 = 32.88grams of Ag
Therefore, 32.88grams of (Ag) silver would plate onto the pure electrode if a current of 6. 8A flowed through the cell for 72 minutes.
Learn more about Ag here;
https://brainly.com/question/3023145
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