2. Refer to Table 10.1, calculate the cell voltage involving: a. Fe (s)/ Fe²+ (0.300M)// Sn²+ (0.150 M)/ Sn (s) b. Mn (s)/ Mn²+ (0.400M)// Ag+ (0.150 M)/ Ag (s) Let me show you an example for a. The cell involves the oxidation of Fe(s) and the reduction of Sn²+, the cell reaction being Fe (s) + Sn²+ → Fe²+ + Sn (s) Excell = Ecathode - Eanode = -0.136-(-0.440) = 0.304 V [Fe²+] log- = 0.304 - [Sn²+] Ecell = E cell 0.0592 2 0.0592 0.300 2 0.500 -log = 0.311 V

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2. Refer to Table 10.1, calculate the cell voltage involving:
a. Fe (s)/ Fe²+ (0.300M)// Sn²+ (0.150 M)/ Sn (s)
b. Mn (s)/ Mn²+ (0.400M)// Ag+ (0.150 M)/ Ag (s)
Let me show you an example for a. The cell involves the oxidation of Fe(s) and the reduction of Sn²+,
the cell reaction being
Ecell Ecathode
Ecell = Eº
cell
Fe (s) + Sn²+ → Fe²+ + Sn (s)
Eanode
[Fe²+]
[Sn²+]
=
0.0592
2
-log-
= -0.136-(-0.440) = 0.304 V
= 0.304 -
0.0592 /0.300
2
0.500,
-log
= 0.311 V
Transcribed Image Text:2. Refer to Table 10.1, calculate the cell voltage involving: a. Fe (s)/ Fe²+ (0.300M)// Sn²+ (0.150 M)/ Sn (s) b. Mn (s)/ Mn²+ (0.400M)// Ag+ (0.150 M)/ Ag (s) Let me show you an example for a. The cell involves the oxidation of Fe(s) and the reduction of Sn²+, the cell reaction being Ecell Ecathode Ecell = Eº cell Fe (s) + Sn²+ → Fe²+ + Sn (s) Eanode [Fe²+] [Sn²+] = 0.0592 2 -log- = -0.136-(-0.440) = 0.304 V = 0.304 - 0.0592 /0.300 2 0.500, -log = 0.311 V
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