Balance the following equations using ion-electrode method. Show all the details. A. Acidic Medium Mno4 + C20, Mn2+ + CO2 Balanced equation: Oxidant: Reductant:
Balance the following equations using ion-electrode method. Show all the details. A. Acidic Medium Mno4 + C20, Mn2+ + CO2 Balanced equation: Oxidant: Reductant:
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.92QE
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Transcribed Image Text:Daro
Mno, + C20,²
Balance the following equations using ion-electrode method. Show all the details.
A. Acidic Medium
Mn2+
CO2
+
Balanced equation:
Oxidant:
Reductant:
![Example A:
CrzO; + C20
CO2
Step 1: Write the half-reactions.
C20
Cr2O;
.CO2
Cra
Step 2: Balance each half-equation "atomically" in the order
a) atoms other than H and O
• 2 CQ
2 Cra
Cr20;
b) O atoms by adding H;0. with the appropriate coefficient, H atoms by adding H- with
the appropriate coefficient
C20
Cr2O 14 H*
→ 2 CO2
→2 Cr + 7 H20
Step 3: Examiné the charge on both sides of the half-reaction. To the more positive side, add
the correct number of.electrons (bearing in mind that each electron has a charge of -1), to equalize
the charge with the other side.
C20
+2 CO2 +
• 2e
Cr20, + 14 H* + 6 e
2 Cr* +7 H20
Step 4: Multiply each entire half-reaction by the lowest factor to equalize the number of
Seiectrons lost and gained.
2 CO2 + 2 e ]
Cr20, + 14 H* + 6 e
→2 Cr + 7 H20 ]
Step 5: Obtain the net redox equation by combining the half-equations. Simplify. Cancel
species, which are found in both sides of the equation, then check by inspection
3 C20
+ 6 CO2 + 6e
Cr2072- + 14 H* + 6 e
2 Cr* + 7 H20
3 C20 + Cr20;2- + 14 H* + 6
6 CO2 + 2 Cr* + 7 H2O + Be
The final balanced equation is:
3 C20,2- + Cr20,2 + 14 H*
6 CO2 +2 Cr* + 7 H20
Check :
reactants
6C
2 Cr
14 H
19 0
products
6C
2 Cr
14 H
19 0
6-
Charge: 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5233f398-8102-4939-a6a5-fbccd1e64ae9%2F96ae34a3-a58f-478c-8787-fa3b706b57f5%2Fsbmfigw_processed.png&w=3840&q=75)
Transcribed Image Text:Example A:
CrzO; + C20
CO2
Step 1: Write the half-reactions.
C20
Cr2O;
.CO2
Cra
Step 2: Balance each half-equation "atomically" in the order
a) atoms other than H and O
• 2 CQ
2 Cra
Cr20;
b) O atoms by adding H;0. with the appropriate coefficient, H atoms by adding H- with
the appropriate coefficient
C20
Cr2O 14 H*
→ 2 CO2
→2 Cr + 7 H20
Step 3: Examiné the charge on both sides of the half-reaction. To the more positive side, add
the correct number of.electrons (bearing in mind that each electron has a charge of -1), to equalize
the charge with the other side.
C20
+2 CO2 +
• 2e
Cr20, + 14 H* + 6 e
2 Cr* +7 H20
Step 4: Multiply each entire half-reaction by the lowest factor to equalize the number of
Seiectrons lost and gained.
2 CO2 + 2 e ]
Cr20, + 14 H* + 6 e
→2 Cr + 7 H20 ]
Step 5: Obtain the net redox equation by combining the half-equations. Simplify. Cancel
species, which are found in both sides of the equation, then check by inspection
3 C20
+ 6 CO2 + 6e
Cr2072- + 14 H* + 6 e
2 Cr* + 7 H20
3 C20 + Cr20;2- + 14 H* + 6
6 CO2 + 2 Cr* + 7 H2O + Be
The final balanced equation is:
3 C20,2- + Cr20,2 + 14 H*
6 CO2 +2 Cr* + 7 H20
Check :
reactants
6C
2 Cr
14 H
19 0
products
6C
2 Cr
14 H
19 0
6-
Charge: 6
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