Question

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E9: Voltaic Cells - Ered of Reduction Half-Cells Purpose Purpose... Results Table 1. Ered of Reduction Half-Cells relative to Cuºl Cu? Voltaic Reduction Half-Cell Cell Cathode Soln. I Cathode Cui Cu +0.309 Ered Rank (1 most +) Ere Rank (1 = most +) [V] NA Ag! | Ag! +0.397 NI- I Niº -0.164 Pb2 Pb -0.462 ZnZnº -1.031 Brz, Br! IC +0.796 L, TIC +0.279 Conclusions (answer the following questions)

fill the table -topic: voltaic cells - Ered of reduction half cells

Answer 1

The oxidation half cell is Cu0|Cu2+.

Eori = Eco Cu2+ = -0.309 V

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Using the relation

Ecell = Ered + Eori

Ered = Ecell + Eom

$E_{Red}=E_{cell}\:-\:0.309\:V$

Voltaic cell	Reduction Half Cell Cathode Soln. | Cathode	ECell [V]	ERed [V]	ERed Rank (1=most+)	E0Red [V]	E0Red Rank (1=most +)
NA	$Cu^{2+}|Cu^{0}$	-	+0.309	2	0.34	4
1	$Ag^{1+}|Ag^{0}$	+0.397	+0.088	3	0.80	2
2	$Ni^{2+}|Ni^{0}$	-0.164	-0.473	5	-0.26	6
3	$Pb^{2+}|Pb^{0}$	-0.462	-0.771	6	-0.13	5
4	$Zn^{2+}|Zn^{0}$	-1.031	-1.340	7	-0.76	7
5	$Br_2,Br^{1-}|C^{0}$	+0.796	0.487	1	1.09	1
6	$I_2,I^{1-}|C^{0}$	+0.279	-0.030	4	0.54	3