Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by = 96487 X 1.45 / 108 C
= 1295.43 C
Given,
Current = 1.5 A
Time =1295.43 /1.5s
= 863.6 s
= 864 s
= 864/ 60
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit = (63.5x1295.43) / (2x96487) g
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit
= (65.4x1295.43) / (2x96487) g
= 0.439 g of Zn
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Further show:
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(iii) Tris(ethane-1,2-diamine) chromium(III) chloride
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(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
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(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec-butyl ketone
(vi) 4-Fluoroacetophenone
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(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
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Thanks
it helped me thanks but not still know the current direction
Very better elaborated than my refresher
Ty for ur help