Redox And Electrochemistry Worksheet With Answer Key Page 6
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8#3 Answer:
2+
→
(a) Sn
Sn
+ 2e- anode reaction
+
→
Ag
+ e-
Ag
+
2+
→
(b) 2 Ag
+ Sn
2 Ag + Sn
E˚ = [0.80v - (-0.14v)] = 0.94v
0.0592
(c) E =
− nFE = −RT ln K
log K OR
n
0.94 × 2
31
log K =
= 31.8 ;
K = 6×10
0.0592
2+
RT
[Sn
]
(d) E = EÞ −
ln Q ; Q =
+
2
nF
[Ag
]
0.0592
1
E = 0.94 −
= 0.84v
log
2
2
(0.02)
#4 Answer:
2−
0.225molS
O
1 mol I
2
3
2
(a) 0.0373L ×
×
=
2−
1L
2 mol S
O
2
3
-3
4.20
10
mol I
×
2
(b) P
= P
- P
H
total
H
O
2
2
= (752 - 24) mm Hg = 728 mm Hg
1 mol H
= 1 mol I
2
2
PV = nRT ; V = nRT/P
(
)
− 3
L⋅atm
(4.20×10
mol) 0.0821
(298K)
mol⋅K
V =
=
728
(
)atm
760
= 0.107 L
- →
(c) At anode: 2 I
I
+ 2e-
2
−
96489amp ⋅ sec
2 mol e
− 3
(d) 4.20 × 10
×
×
×
mol I
−
2
1 mol I
1 mol e
2
1min
1
×
×
= 4.50 amp
60sec.
3min.
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