Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
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). 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. 1.2610^5 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).
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. 1.2610^5 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).
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.
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.
Hi Ans 2: For a steady flow, the current I in amperes can be calculated with the following equation:
Ans 2:
For a steady flow, the current I in amperes can be calculated with the following equation:
where
More generally, electric current can be represented as the time rate of change of charge, or
therefore, Q= 30 A x 70.2 x 60 sec = 126360 coulombs
1 faraday = 96 485.3415 coulombs
Therefore answer is 126360/96485.3415 = 1.31 faraday
Hi Please repost the third question, the question is not clear.
Please repost the third question, the question is not clear.
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
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
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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -