Skip to main content
Chemistry LibreTexts

Balancing Redox Reactions (Solutions)

  • 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}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    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-


    This page titled Balancing Redox Reactions (Solutions) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Draganjac via source content that was edited to the style and standards of the LibreTexts platform.