Given the following reactions: Anode: NO3- + 2H+ + e- --> NO2 + H2O Cathode: 2H+ + 2e- --> H2 2NO2(g) + 2 H2O (I) --> 2NO3-(aq) + 2H+(aq) + Redox: H2(g) Conditions: [0.15 atm] [0.18 atm] [0.1 M] [0.67 atm] Find the pH, if E = - 0.70 V.

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Find the pH, if E = - 0.70 V.
Table 7.2 Stundard potertiabat 298 K. (a) in dectrochemical erder
Reductien half raction
Recuction half-maction
EIV
Strongly onidiring
H + Zir +2J0, + H,0
Se+2S
+0.15
6.07
0,2H+ 0,+HO
S,o +220
+20
00
+2.05
aby definitien
Ce C
In le
Se+ Sn
Au+ Au
-014
N + NI
Ca+ Ce
In+ In
023
+140
Mno,+ H+-+Me"+4H0
Mn" +eMe
+151
Au+ A
a, 2a
Cro + 14H +be +20+7H,0
0,+H0+ 0, 20H
0,+4r 4 20
1.40
In le
+124
123
1.23
-040
-044
Mao, + Ma+21,0
-044
S+
In ln
097
NO; +r NO- 21,0
074
Za+ Zn
-76
+06
NO, +2rr NO, + HO
21,0+ 1, +20r
+0.79
Mn+2e Mn
-LIE
--L19
O+H0+-r +20
+0.76
-163
A Al
Mn +21,0+ 2 Meo,4O
1.79
-209
1,+ 21
+0.54
M+rMg
-2
+0.32
24
252
NICOH + H0-NOI), +Or
NaeNa
-287
+0.00
-2.09
291
Ra a
292
+0.34
-292
-2.93
40.22
2.93
-3.05
Transcribed Image Text:Find the pH, if E = - 0.70 V. Table 7.2 Stundard potertiabat 298 K. (a) in dectrochemical erder Reductien half raction Recuction half-maction EIV Strongly onidiring H + Zir +2J0, + H,0 Se+2S +0.15 6.07 0,2H+ 0,+HO S,o +220 +20 00 +2.05 aby definitien Ce C In le Se+ Sn Au+ Au -014 N + NI Ca+ Ce In+ In 023 +140 Mno,+ H+-+Me"+4H0 Mn" +eMe +151 Au+ A a, 2a Cro + 14H +be +20+7H,0 0,+H0+ 0, 20H 0,+4r 4 20 1.40 In le +124 123 1.23 -040 -044 Mao, + Ma+21,0 -044 S+ In ln 097 NO; +r NO- 21,0 074 Za+ Zn -76 +06 NO, +2rr NO, + HO 21,0+ 1, +20r +0.79 Mn+2e Mn -LIE --L19 O+H0+-r +20 +0.76 -163 A Al Mn +21,0+ 2 Meo,4O 1.79 -209 1,+ 21 +0.54 M+rMg -2 +0.32 24 252 NICOH + H0-NOI), +Or NaeNa -287 +0.00 -2.09 291 Ra a 292 +0.34 -292 -2.93 40.22 2.93 -3.05
Given the following reactions:
Anode: NO3- + 2H+ + e- --> NO2 +
H2O
Cathode: 2H+ + 2e- --> H2
Redox:
2NO2(g) + 2
H2O (I)
--> 2NO3-(aq) + 2H+(aq) +
H2(g)
Conditions:
[0.15 atm]
[0.18 atm]
[0.1 M]
[0.67 atm]
Find the pH, if E = - 0.70 V.
%3D
Transcribed Image Text:Given the following reactions: Anode: NO3- + 2H+ + e- --> NO2 + H2O Cathode: 2H+ + 2e- --> H2 Redox: 2NO2(g) + 2 H2O (I) --> 2NO3-(aq) + 2H+(aq) + H2(g) Conditions: [0.15 atm] [0.18 atm] [0.1 M] [0.67 atm] Find the pH, if E = - 0.70 V. %3D
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