![17. A voltaic cell consists of a Hg/Hg2+ electrode (E°=0.85 V) and a Sn/Sn?* electrode (E°= -0.14 V). Calculate [Sn2] if [Hg2](http://img.homeworklib.com/questions/06815d30-19c1-11ec-a6bf-f9f27d1eecaf.png?x-oss-process=image/resize,w_560)
Homework Answers
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
Answer #1
Ans:- 17.D
18.C
The first problem is solved using Nernst equation.The explanation for both is provided in the picture below.
17 Ans - Given, Hg Hg 2 t = 0,85 v V Sn/542 & E= -0.14 r. I Here for both half celle onidation potential are given. The one with righer oxidation potential will have higher tendency to be onidised and hence behaves and anode. Hehe, reglrg at has higher onidation potential do it behaves as anode and Solat behances as Cathoode. The oxidation potential of Sun 21 = 20.142 so its reduction potential will be = 0.14r. Now, Eley - Exidation & Eseduction = (0.85 +0.14)~ & 0.99 v
et + ae- The half all reactions are anade! &ng - HG+ 2 cathode: Sn2+ + 20 in Overall : 2 Mg + Hodt + Sn. Snat. 02 Applying Nernst equation we have, - 0.059 Thy Ewy - Eccy-n oy Temas] » Putting the giren volnes mee get, 1.04 - 0.99 - 0.059 tag [ont actions transforma 0.05 = – 0.0295 log on 0.48 In = no. L=2 0.48 > -1.69 0.48 101.69 as log x=y a=104 [snet] - 1.69 10 0.48 1.69 x 10 2+ = 0.48 = 23.509 M = 2.35 x 10' & 2.4 El M
(19) Anse Mehan) + 2420 (1) Mnoz (s) + 4HT (an) + de- Giren, cuthent (1) = 760 A. time (t) = 22.0 hours - 22 x 60 x 60 seconds 79 200 seconds .: Cóa ) = 1 x 1 1 1 1 8 = 760 x 79 200 ce = 60, 192000 C Now, 1F = 965 00 c . - F. 965ool 60, 192000 c = 966 -X 60, 192000 f. 96500 623.75 F. from the equation we have I mole of Ano = 2 moles of electrons 2F > IF = moles of Hong » 625 75 f = I x 623.75 moles of Mno = 311.88 moles
NOW, no. of mores - - - Weight (in g) Molar mass .: Mass of ding = no. of modes & Molas mass of Mno, to g/mol of Now, Molar mass of Mono = 86.94 .: Mass of Mno, produced or deposited , = 311.88 X 86.94 = 27,114.41 9 27.114 kg = 2.7 E 1 Kg
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