the electrical cell a and b containing znso4 and CUs o4 solution respectively are connected in series on passing certain amount of electricity 15.2g of zinc was deposited in cell a how much copper would be deposited in cell
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0.426g of copper
Explanation:
Cell B: Ag
+
+e
−
⇌Ag at cathode.
1 mole (108 g) of Ag is deposited by 96500 C.
1.45 g of Ag will be deposited by
108
96500×1.45
=1295.6C.
Now,
Q=It
1295.6=1.5×t
t=864s
Cell A: Zn
2+
+2e
−
→Zn
2 moles of electrons (2×96500 C of current) produces 1 mole (63.5 g) of zinc.
1295.6 C of electricity will deposit
2×96500
65.3
×1295.6=0.438 g of zinc
Cell C: Cu
2+
+2e
−
→Cu
2 moles of electrons (2×96500 C) of current will produce 1 mole (63.5 g) of Cu.
1295.6 C of current will deposit
2×96500
63.5×1295.6
=0.426g of copper
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