In the experiment above, a current was passed for 10 minutes, and 0.63g of copper was found to be deposited on the cathode of CuSO4 cells. The weight of silver deposited on the cathode of AgNO3 cell during the same period would be? [Cu = 63, Ag = 108]
0.54g
1.08g
1.62g
2.16g
3.24g
Explanation
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Discussions (21)
Answer is D.
m1v1/M1=m2v2/M2
Where m1=mass of copper, m2=mass of silver, v1=valency of copper, v2=valency of silver, M1=molar mass of copper, M2=molar mass of silver
0.63*2/63=m2*1/108
m2=108*2*0.63/63=2.16g
Cu mass = 0.63
Cu mm = 63
Ag mm = 108
Ag mass = y?
Recall, nF =Q/mole , Q = mole x nF
Same charge in both so,
mole of Cu x nF = mole of Ag x nF
(0.63/63) x 2(96500) = (y/108) x 1(96500)
0.01 x 2 = y/108
y = 108x0.01x2
y = 2.16g
D is correct ✍️👽
The correct answer is B
mole of copper is 0.63/63=0.01
Since 2F will deposit 1 mole of copper
therefore 0.01 mole ×2=0.02F
1F will deposit 1 mole of silver
0.02F will deposit 1/2×0.02F =0.01 mole of
silver
mole =mass/molar mass
0.01=x/108
mass=108×0.01=108
mind you the question says during the same period it means the same period as copper since we were not given current,dis shows the time given us is of no use, therefore the faraday is going to serve as the period 
A is correct
1 mole of Cu -_------- 2f
1 mole of Ag-----------1f
therefore mole,n=0.63/63=0.01
so, 0.01 of Cu-----------0.02f
=> 0.005 of Ag =0.01 of Cu
mass = n xmm= 0.005 x 108 =0.54g
