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1. Exactly 0.200 mole electrons is passed through two electrolytic cells in series. One contains silver ion, and the other, zinc ion .assume that the only cathode reaction in each cell is the reduction of the ion to the metal. How many grams of each metal will be deposited? (at. Mass of zn = 65, ag = 108) a. 108g ag and 65 g zn. b. 108 ag and 32.5g zn. c. 21.6 ag and 6.5 g zn. d. 21.6 g ag and 13 g zn. 2. A steady current of 30.0A for 70.2 min corresponds to a passage of a. 1.26 10^5 electrons. b. 1.26×10^5 f. c. 1.31 f. d. 1.31 c. 3. Cu+ is not stable and undergoes disproportion. E for cu+ disproportionation. (E ° cu2+/cu =0.153V , E ° cu+/cu=0.53V).


11 years ago

							Hi
Ans 1:
For silver , the conversion is Ag+ + e- -> Ag
Therefore 1 mole of electrons are reqd to make 1 mole i.e 108 gm of silver. hence .2 moles of electron will make .2x108 = 21.6 gm of silver.

For zinc, the connversion is Zn2+ + 2e- -> Zn
Therefore 2 moles of elctrons are reqd to make 1 mole i.e 65 gm of zinc, hence .2 mole electrons shall make .2x65/2 = 6.5 gm of zinc.

11 years ago
							Hi
Ans 2:
For a steady flow, the current I in amperes can be calculated with the following equation: $I = {Q \over t}$
where
Q is the electric charge in coulombs transferred t is the time in seconds
More generally, electric current can be represented as the time rate of change of charge, or $I = \frac{dQ}{dt}$.
therefore, Q= 30 A x 70.2 x 60 sec = 126360 coulombs
1 faraday = 96 485.3415 coulombs


11 years ago
							Hi
Please repost the third question, the question is not clear.

11 years ago
							1)       for silver deposition :       Ag+     +     e-             ------>   Ag
1 mole                   1 mole
so , 0.2 moles of Ag is deposited i.e., 0.2*108 = 21.6 grams

for zinc deposition :   Zn2+     +   2e-                ------>    Zn
2 mole                       1 mole
so , 0.1 moles of Zn is deposited i.e., 0.1*65 =  6.5 grams
So option is C

2)   Electric current is the rate of charge flow past a given point in an electric circuit, measured in Coulombs/second which is named Amperes
so Q = i*t
Q=30*70.2*60
Q = 126360 Coulombs
but 1 faraday = 96485 Coulombs
So its Q = 1.31 faraday
So option is C

3) I want to make a small correction to question reg. elecric potential of  cu2+/cu+ ( taken frm standard books )

Its cu2+/cu+ =0.153V  i.e.,                     cu2+ + e-   ------>  cu+            E0 = 0.153 V

Now going into problem:  Cu+ solutions, tend to give a disproportion reaction:

2Cu+ ----- >   Cu + Cu2+

cu+ + e-    ------>  cu                                E0 = 0.53 V        so,  nE0 = 0.53 V

cu+            ------>  cu2+ + e-               E0 = - 0.153 V       so,   nE0 = - 0.153 V

on adding above rxns we get
2Cu+ ----- >   Cu + Cu2+                  with  E0 = +0.377 V

11 years ago
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