BALANCE REDOX REACTION (USE THE LONG METHOD WITH CHECKINGS) MnO2 + CrO3 → Mn + CrO4 - CONTENT SHOULD BE THESE OXIDIZED REDUCED BASIC SOLUTION CHECKING 1 3) Al P+ Al V oxidize reducing Oxidized: A AL- E Al 2e AL+

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Chapter18: Electrochemistry
Section: Chapter Questions
Problem 156IP: An electrochemical cell is set up using the following unbalanced reaction: Ma+(aq)+N(s)N2+(aq)+M(s)...
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FOLLOW THE METHOD FROM THE OTHER PHOTO PLEASE

BALANCE REDOX REACTION
(USE THE LONG METHOD WITH
CHECKINGS)
MnO₂ + CrO3¯¯ → Mn + CrO4¯¯
CONTENT SHOULD BE THESE
OXIDIZED
REDUCED
BASIC SOLUTION
CHECKING
2+
3) AI + NO₂ → A1²+ + ND ₂
+4 -4 =0
↑
+42
+2
+5-2
Al + NO ₂ → A1²+ + NO₂
+5+₂=1
↑
oxidized, reduced, oridizing agent
reducing agent
oxidized:
A1→ 41²+ + 2₂
+2-2=0
ducking AI:1
#:4
U2
0:6
AL: 1 141:1
0
Al212e
2e + 4H+ + 2NO → 2ND ₂ + 2H₂P
2
AL + 2 + 4H+ + 2NO₂ → A1 ²+ + ²√²+ 2NO₂ + 2H₂O
| Al + 4H+ + 2 NO ₂² -> A1 ²+ + 2ND ₂ + 2H₂O Balanted in acidic solution
+4-2=+2
BASIC SOLUTION
Al: T
H÷4
N÷2
0216=8
V=2
-2
0:4+2=6
+2
Reduced:
et 2H+ + NO₂ → NO ₂ + H₂D
V: 1
N = 1
0.83
H: 2
O
Date :
0:3
H.2
+2 H-EO
#:4
0² 4+4=8
N÷2
+2-4 = -2v
2e
2 (e¯ + 2H+ + NO₂¯ →→ NO ₂ + H₂O)
2e + 4H+ + 2ND₂ → 2NO₂ + 2H₂O
AL + 4H+ + 2NO3 → A1 ²+ + 2NO₂ + 2H₂O
Al + 4H+ + 40H- + 2 NO ₂- → AF²+ + 2NO₂ + 2H₂0 + 40H-
Al + ₂ + 2 N0₂- →
A1 ²+ + 2NO₂ + 2H₂Q + 40H-
Al + 2 H₂O + 2NO3 → A1²+ + 2NO₂ + 40H - Balanced in basic solution
checking
VICTORY
Transcribed Image Text:BALANCE REDOX REACTION (USE THE LONG METHOD WITH CHECKINGS) MnO₂ + CrO3¯¯ → Mn + CrO4¯¯ CONTENT SHOULD BE THESE OXIDIZED REDUCED BASIC SOLUTION CHECKING 2+ 3) AI + NO₂ → A1²+ + ND ₂ +4 -4 =0 ↑ +42 +2 +5-2 Al + NO ₂ → A1²+ + NO₂ +5+₂=1 ↑ oxidized, reduced, oridizing agent reducing agent oxidized: A1→ 41²+ + 2₂ +2-2=0 ducking AI:1 #:4 U2 0:6 AL: 1 141:1 0 Al212e 2e + 4H+ + 2NO → 2ND ₂ + 2H₂P 2 AL + 2 + 4H+ + 2NO₂ → A1 ²+ + ²√²+ 2NO₂ + 2H₂O | Al + 4H+ + 2 NO ₂² -> A1 ²+ + 2ND ₂ + 2H₂O Balanted in acidic solution +4-2=+2 BASIC SOLUTION Al: T H÷4 N÷2 0216=8 V=2 -2 0:4+2=6 +2 Reduced: et 2H+ + NO₂ → NO ₂ + H₂D V: 1 N = 1 0.83 H: 2 O Date : 0:3 H.2 +2 H-EO #:4 0² 4+4=8 N÷2 +2-4 = -2v 2e 2 (e¯ + 2H+ + NO₂¯ →→ NO ₂ + H₂O) 2e + 4H+ + 2ND₂ → 2NO₂ + 2H₂O AL + 4H+ + 2NO3 → A1 ²+ + 2NO₂ + 2H₂O Al + 4H+ + 40H- + 2 NO ₂- → AF²+ + 2NO₂ + 2H₂0 + 40H- Al + ₂ + 2 N0₂- → A1 ²+ + 2NO₂ + 2H₂Q + 40H- Al + 2 H₂O + 2NO3 → A1²+ + 2NO₂ + 40H - Balanced in basic solution checking VICTORY
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