Question

1) Making use of Table 1, write out individual redox half reactions (and their balanced sum, including overall potential) that describe the net chemical reactions in procedures (A-F).

A
Complexation:
CuCl²⁺ + 4NH₃ --> [Cu(NH₃)₄]Cl₂
Net Ionic:
Cu²⁺ + 4NH₃ --> Cu(NH₃)₄⁺

B
Complexation:
CuCl₂ + 2en --> [Cu(en)₂]Cl₂
Net Ionic:
Cu²⁺ + 2en --> Cu(en)²⁺

C
Reduction:
2CuCl₂ + 3NH₄Cl --> NH₄[CuCl₃] + (NH₄)₂[CuCl₄]
Net Ionic:
Cu²⁺ + 2Cl^- + Cu --> 2CuCl

D
Reduction:
2CuCl₂ + 4KI --> 2CuI + 4KCl + I₂
Net Ionic:
2Cu²⁺ + 4I^- --> 2CuI + I₂

E
Oxidation:
2CuI + 8NH₃ --> [Cu(NH₃)₄]₂ + I2
Net Ionic:
Cu⁺ + 4NH₃ --> Cu(NH₃)₄²⁺

F
Oxidation:
2CuI + 4en --> [Cu(en)₂]₂ + I₂
Net Ionic:
Cu(en)₂⁺ --> Cu(en)₂²⁺ + Cu + 2en

Table 1

	Couple			E $\degree$ value
(1)	Cu2+ + I- + e-	$\leftrightharpoons$	CuI	0.86 V
(2)	Cu2+ + Cl- + e-	$\leftrightharpoons$	CuCl	0.54 V
(3)	I2 + 2e-	$\leftrightharpoons$	2I-	0.54 V
(4)	Cu+ (aq) + e-	$\leftrightharpoons$	Cu(s)	0.52 V
(5)	Cu2+ (aq) + 2e-	$\leftrightharpoons$	Cu(s)	0.37 V
(6)	CuCl + e-	$\leftrightharpoons$	Cu + Cl-	0.14 V
(7)	Cu(NH3)42+ + 2e-	$\leftrightharpoons$	Cu + 4NH3	-0.12 V
(8)	Cu2+ (aq) + e-	$\leftrightharpoons$	Cu+ (aq)	-0.15 V
(9)	CuI + e-	$\leftrightharpoons$	Cu + I-	-0.19 V
(10)	Cu(en)22+ + 2e-	$\leftrightharpoons$	Cu + 2en	-0.50 V

[en = ethylenediamine]

Answer 1