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. %3D

Chemistry: Principles and Reactions
8th Edition
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Publisher:William L. Masterton, Cecile N. Hurley
Chapter16: Spontaneity Of Reaction
Section: Chapter Questions
Problem 72QAP: Given the following standard free energies at 25°C for the following reactions:...
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Given the following reactions:
Anode: NO3- + 2H+ + e- --> NO2 +
H2O
Cathode: 2H+ + 2e- --> H2
2NO2(g) + 2
H2O (1) --> 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.
Transcribed Image Text:Given the following reactions: Anode: NO3- + 2H+ + e- --> NO2 + H2O Cathode: 2H+ + 2e- --> H2 2NO2(g) + 2 H2O (1) --> 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.
Find the pH, if E = - 0.70 V.
Table 7.2 Standard potentialsat 298 K. (a) In clectrochemical order
Reduction half raction
Reduction half-reaction
EIV
Cu"+eCu
Se+ 2 Se
Agie +e Ag+ Br
Strongly oxidizing
+0.16
HXeO, + 2H + 2e → XeO, + 3H,0
+30
+0.15
2.87
+0.07
0,+ 2H + 0,+H,0
S,0 +2e +2S0
+207
0.00
2H"+ H,
by definition
+2.05
1.98
-0.04
0.06
H,0, + 2H+ 2- 21,0
+1.78
In' ln
Se+e Sn
Au" +e Au
+1.69
-0.14
Pb + 2e P
1.67
0.14
ZHCIO + 2H + 2e-a,+21,0
Agl g+r
N+2eNi
+1.63
-0.15
+161
023
2BrO+2r + 2e , + 21,0
Co+2e Ce
In+*In
+1.60
Mn0; + H" + S+Me" + 4H,0
Mn" +eMn
+1.51
-0.34
+1.51
-034
Au+ Au
+1.40
PASO,+ 2b+ so
+1.36
-0.37
Cr0 + 14H +6e →20" +7H,0
0,+H0++0, + 20H
0,+ 4H* + 4 2H,0
1.33
-.40
+1.24
Im ln
--0.40
1.23
1.23
Fe+ e
-0.44
Mno,+4+ 2 Ma" + 2H,0
+1.23
In+ In
-0.44
+1.09
-0.48
+0.97
In +eIn"
-49
NO, ++ NO+ 21,0
2Hg" + 2 g
CIo +H,0+2 -d+20
Hig +2e -Hg
NO, +2H +- NO, +HO
0.6
061
0.92
-074
Za + Zn
-0.76
+0.6
CatOH), + 2Cd+ 20H
21,0+ 2e1, +20r
C+2 Cr
Mn+2e Mn
+0.80
+0.80
-091
Hg + 22g
+0.79
-LIE
+0.77
-L19
BrO + H,0+ 2r+ir + 20H
FigS0, + 2e+2ig+ so
Mno + 211,0+ 2 Meo, + 4O
Mno, +e Mno
1+ 21
+0.76
-163
062
A Al
-1.66
U+e U
1.79
+0.56
S+eSc
-209
Mg+ 2Mg
Ce+ Ce
+0.54
-236
+0.52
-2.48
L la
Na +e Na
+0.53
-252
NICOH + H0e Ni(OH), + OH
AgCro, + 2e2Ag+ Co
0,+2H,0+e4OH
do, +H0+2 ao, + 20H
0.49
-271
+0.45
C +2Ca
-287
40.40
-289
+0.36
-2.91
0.36
Ra+2 Ra
-2.92
C+ 2eCu
+034
-2.92
Hga, + 2+2Hg+2
40.27
Rb +eRb
-2.93
40.22
2.93
+0.20
-305
Transcribed Image Text:Find the pH, if E = - 0.70 V. Table 7.2 Standard potentialsat 298 K. (a) In clectrochemical order Reduction half raction Reduction half-reaction EIV Cu"+eCu Se+ 2 Se Agie +e Ag+ Br Strongly oxidizing +0.16 HXeO, + 2H + 2e → XeO, + 3H,0 +30 +0.15 2.87 +0.07 0,+ 2H + 0,+H,0 S,0 +2e +2S0 +207 0.00 2H"+ H, by definition +2.05 1.98 -0.04 0.06 H,0, + 2H+ 2- 21,0 +1.78 In' ln Se+e Sn Au" +e Au +1.69 -0.14 Pb + 2e P 1.67 0.14 ZHCIO + 2H + 2e-a,+21,0 Agl g+r N+2eNi +1.63 -0.15 +161 023 2BrO+2r + 2e , + 21,0 Co+2e Ce In+*In +1.60 Mn0; + H" + S+Me" + 4H,0 Mn" +eMn +1.51 -0.34 +1.51 -034 Au+ Au +1.40 PASO,+ 2b+ so +1.36 -0.37 Cr0 + 14H +6e →20" +7H,0 0,+H0++0, + 20H 0,+ 4H* + 4 2H,0 1.33 -.40 +1.24 Im ln --0.40 1.23 1.23 Fe+ e -0.44 Mno,+4+ 2 Ma" + 2H,0 +1.23 In+ In -0.44 +1.09 -0.48 +0.97 In +eIn" -49 NO, ++ NO+ 21,0 2Hg" + 2 g CIo +H,0+2 -d+20 Hig +2e -Hg NO, +2H +- NO, +HO 0.6 061 0.92 -074 Za + Zn -0.76 +0.6 CatOH), + 2Cd+ 20H 21,0+ 2e1, +20r C+2 Cr Mn+2e Mn +0.80 +0.80 -091 Hg + 22g +0.79 -LIE +0.77 -L19 BrO + H,0+ 2r+ir + 20H FigS0, + 2e+2ig+ so Mno + 211,0+ 2 Meo, + 4O Mno, +e Mno 1+ 21 +0.76 -163 062 A Al -1.66 U+e U 1.79 +0.56 S+eSc -209 Mg+ 2Mg Ce+ Ce +0.54 -236 +0.52 -2.48 L la Na +e Na +0.53 -252 NICOH + H0e Ni(OH), + OH AgCro, + 2e2Ag+ Co 0,+2H,0+e4OH do, +H0+2 ao, + 20H 0.49 -271 +0.45 C +2Ca -287 40.40 -289 +0.36 -2.91 0.36 Ra+2 Ra -2.92 C+ 2eCu +034 -2.92 Hga, + 2+2Hg+2 40.27 Rb +eRb -2.93 40.22 2.93 +0.20 -305
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