Balancing Redox Reactions (Solutions)

Last updated
Save as PDF

Page ID: 11049

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

1.

Bi(OH)₃ + SnO₂²^- ---------> SnO₃²^- + Bi (basic solution)

Step 1. Break into half-reactions:

Bi(OH)₃ -------> Bi
SnO₂²^- -------> SnO₃²^-

Step 2. Balance atoms other than H and O

Bi(OH)₃ -------> Bi
SnO₂²^- -------> SnO₃²^-

Step 3. Balance O by adding H₂O

Bi(OH)₃ -------> Bi + 3 H₂O
H₂O + SnO₂²^- -------> SnO₃²^-

Step 4. Balance H by adding H⁺

3 H⁺ + Bi(OH)₃ -------> Bi + 3 H₂O
H₂O + SnO₂²^- -------> SnO₃²^-+ 2H⁺

Step 5. Balance charge by adding electron(s)

3 e^- + 3 H⁺ + Bi(OH)₃ -------> Bi + 3 H₂O
H₂O + SnO₂²^- -------> SnO₃²^-+ 2H⁺ + 2 e^-

Step 6. Electrons lost = electrons gained

(3 e^- + 3 H⁺ + Bi(OH)₃ -------> Bi + 3 H₂O) x 2
( H₂O + SnO₂²^- -------> SnO₃²^-+ 2H⁺ + 2 e^- ) x 3

Gives:
6 e^- + 6 H⁺ + 2 Bi(OH)₃ -------> 2 Bi + 6 H₂O
3 H₂O + 3 SnO₂²^- -------> 3 SnO₃²^-+ 6H⁺ + 6 e^-

Step 7. Cancel like terms and add half reactions

2 Bi(OH)₃ + 3 SnO₂²^- -------> 2Bi + 3 SnO₃²^-+ 3 H₂O

Step 8. Since it is a basic solution, add OH^- for each H⁺ (add OH^- to both sides of eqn)

No H⁺, finished

2.

S₂O₃²^- + I₂ --------> I^- + S₄O₆²^- (acidic solution)

Step 1. Break into half-reactions:

S₂O₃²^- -------> S₄O₆²^-
I₂ -------> I^-

Step 2. Balance atoms other than H and O

2 S₂O₃²^- -------> S₄O₆²^-
I₂ -------> 2 I^-

Step 3. Balance O by adding H₂O (already balanced)

2 S₂O₃²^- -------> S₄O₆²^-
I₂ -------> 2 I^-

Step 4. Balance H by adding H⁺(No H's)

2 S₂O₃²^- -------> S₄O₆²^-
I₂ -------> 2 I^-

Step 5. Balance charge by adding electron(s)

2 S₂O₃²^- -------> S₄O₆²^- + 2 e^-
2 e^- + I₂ -------> 2 I^-

Step 6. Electrons lost = electrons gained (2 e^- lost, 2 e^- gained)

2 S₂O₃²^- -------> S₄O₆²^- + 2 e^-
2 e^- + I₂ -------> 2 I^-

Step 7. Cancel like terms and add half reactions

2 S₂O₃²^- + I₂ -------> S₄O₆²^- + 2 I^-

3

MnO₄^- + I^- --------> MnO₂ + I₂ (basic solution)

Step 1. Break into half-reactions:

MnO₄^- -------> MnO₂
I^- -------> I₂

Step 2. Balance atoms other than H and O

MnO₄^- -------> MnO₂
2 I^- -------> I₂

Step 3. Balance O by adding H₂O

MnO₄^- -------> MnO₂ + 2 H₂O
2 I^- -------> I₂

Step 4. Balance H by adding H⁺

4 H⁺ + MnO₄^- -------> MnO₂ + 2 H₂O
2 I^- -------> I₂

Step 5. Balance charge by adding electron(s)

3 e^- + 4 H⁺ + MnO₄^- -------> MnO₂ + 2 H₂O
2 I^- -------> I₂ + 2 e^-

Step 6. Electrons lost = electrons gained

(3 e^- + 4 H⁺ + MnO₄^- -------> MnO₂ + 2 H₂O) x 2
( 2 I^- -------> I₂ + 2 e^- ) x 3

Gives:
6 e^- + 8 H⁺ + 2 MnO₄^- -------> 2 MnO₂ + 4 H₂O
6 I^- -------> 3 I₂ + 6 e^-

Step 7. Cancel like terms and add half reactions

8 H⁺ + 2 MnO₄^- + 6 I^- -------> 2 MnO₂ + 4 H₂O + 3 I₂

Step 8. Since it is a basic solution, add OH^- for each H⁺ (add OH^- to both sides of eqn)

8 OH^- + 8 H⁺ + 2 MnO₄^- + 6 I^- -------> 2 MnO₂ + 4 H₂O + 3 I₂ + 8 OH^-

9. Combine H⁺ with OH^- to form H₂O

8 OH^- + 8 H⁺ + 2 MnO₄^- + 6 I^- -------> 2 MnO₂ + 4 H₂O + 3 I₂ + 8 OH^-
Gives:
8 H₂O + 2 MnO₄^- + 6 I^- -------> 2 MnO₂ + 4 H₂O + 3 I₂ + 8 OH^-

10. Cancel like terms:

4 H₂O + 2 MnO₄^- + 6 I^- -------> 2 MnO₂ + 3 I₂ + 8 OH^-