Coding the Future

Given I Cu2 2e в Cu Eв 0 337 V Ii Cu2 в в Cut Eв 0

3 given 1 cu2 2e cu E0 0 337v 2 cu2 e Cu E0 0 153v
3 given 1 cu2 2e cu E0 0 337v 2 cu2 e Cu E0 0 153v

3 Given 1 Cu2 2e Cu E0 0 337v 2 Cu2 E Cu E0 0 153v 57. given (i) cu2 2e– cu, eo = 0.337 v (ii) cu2 e– cu , eo = 0.153 v electrode potential, eo for the reaction, cu e– cu, will be (1) 0.38 v (2) 0.52 v (3) 0.90 v (4) 0.30 v. Given : (i) cu2 2e → cu, e° = 0.337 v (ii) cu2 e → cu , e° = 0.153 v electrode potential, e° v (b) 0.30 v (c) 0.38 v (d) 0.52 v.

юааgivenюаб I юааcuюаб 2 юаа2eюаб тжт юааcuюаб юааeюаб тиш юаа0юаб юаа337юаб юааvюаб юааiiюаб юааcuю
юааgivenюаб I юааcuюаб 2 юаа2eюаб тжт юааcuюаб юааeюаб тиш юаа0юаб юаа337юаб юааvюаб юааiiюаб юааcuю

юааgivenюаб I юааcuюаб 2 юаа2eюаб тжт юааcuюаб юааeюаб тиш юаа0юаб юаа337юаб юааvюаб юааiiюаб юааcuю 147 views. asked oct 10, 2022 in chemistry by deepika tiwari (56.9k points) given. i) cu2 2 e → cu, e° = 0.337. ii) cu2 e → cu , e° = 0.153 v. electrode potential, e for the reaction, 2 . cu2 e → cu, will be. a) 0.90 v. b) 0.30 v. 57. given (i) cu2 2e– cu, eo = 0.337 v (ii) cu2 e– cu , eo = 0.153 v electrode potential, eo for the reaction, cu e– cu, will be (1) 0.38 v (2) 0.52 v (3) 0.90 v (4) 0.30 v. 0.52 v. explanation of the correct answer: the correct answer is option c: from the given, (1) c u 2 2 e → c u e ° = 0. 337 v Δ g = n f e ° ⇒ Δ g = 2 * f * 0. 3370 [i] (2) c u 2 e → c u e ° = 0. 153 v Δ g = n f e ° ⇒ Δ g = 1 * f * 0. 153 [i i] adding the equations (i) and (ii), c u e → c u Δ g ° = 0. 521 f Δ g. The electrode potential for the reaction is e∘=0.521 v option (d) is correct. explanation: Δg∘ = nfe∘ for, cu2 2e− → cu , Δg∘1 = −2f ( 0.337 ) . . . . (1) cu2 e− → cu , Δg∘2 = −f(0.153) . . . (2) subtracting (2) from (1), cu e− → cu Δg∘ = −0.674 f 0.153 f . . . (3) nfe∘ = − 0.521 f for equation.

57 given i Cu2 2eвђ cu Eo 0 337 v ii cu2 eо
57 given i Cu2 2eвђ cu Eo 0 337 v ii cu2 eо

57 Given I Cu2 2eвђ Cu Eo 0 337 V Ii Cu2 Eо 0.52 v. explanation of the correct answer: the correct answer is option c: from the given, (1) c u 2 2 e → c u e ° = 0. 337 v Δ g = n f e ° ⇒ Δ g = 2 * f * 0. 3370 [i] (2) c u 2 e → c u e ° = 0. 153 v Δ g = n f e ° ⇒ Δ g = 1 * f * 0. 153 [i i] adding the equations (i) and (ii), c u e → c u Δ g ° = 0. 521 f Δ g. The electrode potential for the reaction is e∘=0.521 v option (d) is correct. explanation: Δg∘ = nfe∘ for, cu2 2e− → cu , Δg∘1 = −2f ( 0.337 ) . . . . (1) cu2 e− → cu , Δg∘2 = −f(0.153) . . . (2) subtracting (2) from (1), cu e− → cu Δg∘ = −0.674 f 0.153 f . . . (3) nfe∘ = − 0.521 f for equation. Given:(i) cu2 2e−→cu,e0=0.337 v(ii) cu2 e−→cu ,e0=0.153standard electrode potential, e∘ for the reaction, cu e−→cu, will be (in volts):. Neet jee mcq 4hello myself ayon das in this video i discusgiven, (i) cu2 2e → cu, eo = 0.337 v(ii) cu2 e → cu , eo = 0.153electrode potential,.

31 given i Cu2 2e в cu eв 0 337 v ii cu2
31 given i Cu2 2e в cu eв 0 337 v ii cu2

31 Given I Cu2 2e в Cu Eв 0 337 V Ii Cu2 Given:(i) cu2 2e−→cu,e0=0.337 v(ii) cu2 e−→cu ,e0=0.153standard electrode potential, e∘ for the reaction, cu e−→cu, will be (in volts):. Neet jee mcq 4hello myself ayon das in this video i discusgiven, (i) cu2 2e → cu, eo = 0.337 v(ii) cu2 e → cu , eo = 0.153electrode potential,.

163 given cu 2 2e cu e O 0 32 v Ksp Of cu Oh 2 1
163 given cu 2 2e cu e O 0 32 v Ksp Of cu Oh 2 1

163 Given Cu 2 2e Cu E O 0 32 V Ksp Of Cu Oh 2 1

Comments are closed.