Balancing Redox Equations and Half-Reactions Practice Questions — JAMB UTME

Question 1Question →

Complete the reduction half-reaction by identifying the correct coefficient for the electrons needed to balance the charge.

Fill in the blanks below

In the balanced reduction half-reaction in acidic medium: $\text{MnO}_4^- + 8\text{H}^+ + e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, the missing coefficient for the electrons is .

Show answer & explanation

Answer

5

The total charge on the left side of the equation is

+7

(from one

\text{MnO}_4^-

ion carrying

-1

and eight

\text{H}^+

ions carrying

+8

). The total charge on the right side is

+2

(from one

\text{Mn}^{2+}

ion). To make both sides equal in charge, 5 electrons (each having a

-1

charge) must be added to the reactant side.

Step-by-Step Solution

1

Calculate the total ionic charge on the reactant side before adding electrons.

Net reactant charge =

(-1) + 8(+1) = +7

One permanganate ion contributes a charge of

-1

and eight hydrogen ions contribute

+8

.

2

Calculate the total ionic charge on the product side.

Net product charge =

+2

One manganese(II) ion contributes

+2

and four water molecules are neutral (

0

).

3

Determine the number of electrons required to balance the overall charge.

+7 + 5(-1) = +2

, so 5 electrons are required.

Electrons carry a

-1

charge, so adding

5e^-

to the reactant side lowers its charge from

+7

to

+2

to match the product side.

Key Concept

Balancing charge in half-reactions by adding electrons

Question 2Question →

When the following redox reaction is balanced in an acidic medium using the smallest whole-number coefficients:

\text{ClO}_3^-(\text{aq}) + a\text{Fe}^{2+}(\text{aq}) + b\text{H}^+(\text{aq}) \rightarrow \text{Cl}^-(\text{aq}) + c\text{Fe}^{3+}(\text{aq}) + d\text{H}_2\text{O}(\text{l})

What is the value of the coefficient

b

for

\text{H}^+

?

3

5

6

12

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient for hydrogen ions is 6.

In the reduction half-reaction,

\text{ClO}_3^-

is reduced to

\text{Cl}^-

. Balancing the 3 oxygen atoms requires

3\text{ H}_2\text{O}

on the product side. Consequently,

6\text{ H}^+

ions are required on the reactant side to balance the 6 hydrogen atoms, giving

b = 6

.

Step-by-Step Solution

1

Determine the oxidation state change for chlorine

In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

. In

\text{Cl}^-

, chlorine has an oxidation state of

-1

. The total change is a gain of

6\text{ e}^-

.

Knowing the electron transfer per mole of chlorate ion establishes the electron requirement for the reduction half-reaction.

2

Balance oxygen atoms using water

The chlorate ion

\text{ClO}_3^-

contains 3 oxygen atoms, requiring

3\text{ H}_2\text{O}

on the product side.

In acidic redox balancing, oxygen atoms are balanced by adding water molecules to the side deficient in oxygen.

3

Balance hydrogen atoms using hydrogen ions

To balance the 6 hydrogen atoms in

3\text{ H}_2\text{O}

, add

6\text{ H}^+

to the reactant side.

Hydrogen atoms from the water molecules on the product side must originate from hydrogen ions in the acidic medium.

4

Verify overall mass and charge balance

The reduction half-reaction is

\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}

. Adding the oxidation half-reaction

6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-

yields the overall balanced equation with

b = 6

.

Combining half-reactions confirms that electrons cancel out and mass and charge are conserved.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Question 3Question →

Consider the redox reaction between dichromate ions (

\text{Cr}_2\text{O}_7^{2-}

) and iron(II) ions (

\text{Fe}^{2+}

) in an acidic medium:

\text{Cr}_2\text{O}_7^{2-} + x\text{Fe}^{2+} + y\text{H}^+ \rightarrow 2\text{Cr}^{3+} + x\text{Fe}^{3+} + z\text{H}_2\text{O}

What is the stoichiometric coefficient

x

of

\text{Fe}^{2+}

when the ionic equation is completely balanced?

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient x of Fe²⁺ in the balanced redox reaction is 6.

In the reduction half-reaction, dichromate (

\text{Cr}_2\text{O}_7^{2-}

) contains two Cr atoms in the +6 oxidation state converting to two

\text{Cr}^{3+}

ions in the +3 state, which consumes 6 electrons. In the oxidation half-reaction, each

\text{Fe}^{2+}

ion loses 1 electron to form

\text{Fe}^{3+}

. To balance charge transfer, 6

\text{Fe}^{2+}

ions are needed for every 1

\text{Cr}_2\text{O}_7^{2-}

ion, making the stoichiometric coefficient

x

equal to 6.

Step-by-Step Solution

1

Determine the oxidation state changes for Chromium and Iron.

Chromium changes from +6 in

\text{Cr}_2\text{O}_7^{2-}

to +3 in

\text{Cr}^{3+}

, requiring 3 electrons per Chromium atom (

6e^-

total for two Cr atoms). Iron changes from +2 in

\text{Fe}^{2+}

to +3 in

\text{Fe}^{3+}

, releasing

1e^-

per Iron atom.

Identifying the number of electrons transferred in each half-reaction is required to balance the overall redox equation.

2

Balance the electron gain and loss.

The oxidation half-reaction (

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

) must be multiplied by 6 to balance the 6 electrons required by the dichromate ion.

The total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent.

3

Combine the half-reactions and read the coefficient

x

.

The balanced chemical equation is

\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}

, giving

x = 6

.

The coefficient

x

corresponds directly to the stoichiometric multiplier applied to

\text{Fe}^{2+}

.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Question 4Question →

Consider the following ionic equation for the redox reaction between dichromate(VI) ions and iodide ions in an acidic medium:

\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + x\text{I}^-(\text{aq}) + y\text{H}^+(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + z\text{I}_2(\text{aq}) + w\text{H}_2\text{O}(\text{l})

What are the correct values of the stoichiometric coefficients

x

,

y

, and

z

respectively when the equation is completely balanced?

6, 14, 3

2, 14, 1

6, 7, 3

3, 14, 6

Show answer & explanation

Answer: 6, 14, 3

Answer

The correct stoichiometric coefficients for x, y, and z are 6, 14, and 3 respectively.

The reduction of one dichromate ion (

\text{Cr}_2\text{O}_7^{2-}

) requires 6 electrons and 14 hydrogen ions to yield two

\text{Cr}^{3+}

ions and 7 water molecules. Oxidation of iodide ions (

\text{I}^-

) to iodine (

\text{I}_2

) releases 2 electrons per molecule formed. To equalize electron exchange at 6 electrons, 6 moles of iodide ions (

x = 6

) produce 3 moles of iodine molecules (

z = 3

), requiring 14 moles of

\text{H}^+

(

y = 14

).

Step-by-Step Solution

1

Write and balance the reduction half-reaction for dichromate(VI) ions

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Chromium changes oxidation state from +6 to +3 (total 6 electrons gained for two Cr atoms), and 7 oxygen atoms require

14\text{H}^+

to form

7\text{H}_2\text{O}

.

2

Write and balance the oxidation half-reaction for iodide ions

2\text{I}^- \rightarrow \text{I}_2 + 2e^-

Iodine changes oxidation state from -1 to 0, losing 1 electron per iodide ion.

3

Equalize the number of electrons transferred in both half-reactions

Multiply the oxidation half-reaction by 3:

6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-

6 electrons lost must equal 6 electrons gained.

4

Combine the half-reactions and determine coefficients

\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}

Matching coefficients gives

x = 6

,

y = 14

, and

z = 3

.

Key Concept

Balancing Redox Equations using Half-Reactions in Acidic Medium

Question 5Question →

Complete the balanced reduction half-reaction equation for the conversion of nitrate ions to nitrogen monoxide gas in an acidic medium by providing the missing stoichiometric coefficients. What are the values of the coefficients for hydrogen ions and electrons?

Fill in the blanks below

$$\text{NO}_3^-(\text{aq}) + \text{H}^+(\text{aq}) + e^- \rightarrow \text{NO}(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$

Show answer & explanation

Answer

The balanced reduction half-reaction requires 4 hydrogen ions (H⁺) and 3 electrons (e⁻).

The balanced half-reaction is NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l). Four hydrogen ions balance the four hydrogens present in the two water molecules on the right. Three electrons balance the net charge (+3 on the left reactant side versus 0 on the neutral product side), confirming that nitrogen is reduced from oxidation state +5 in NO₃⁻ to +2 in NO.

Step-by-Step Solution

1

Balance oxygen atoms using water molecules

One nitrate ion (NO₃⁻) has 3 oxygen atoms, and NO has 1 oxygen atom, requiring 2 H₂O molecules on the product side: NO₃⁻ → NO + 2 H₂O

In aqueous acid media, oxygen atoms are balanced by adding H₂O molecules to the oxygen-deficient side.

2

Balance hydrogen atoms using hydrogen ions

The right side has 4 hydrogen atoms in 2 H₂O, so 4 H⁺ ions must be added to the left side: NO₃⁻ + 4 H⁺ → NO + 2 H₂O

Hydrogen atoms in acidic media are balanced by adding H⁺ ions to the hydrogen-deficient side.

3

Balance electrical charge using electrons

Left side net charge = (-1) + 4(+1) = +3. Right side net charge = 0. Adding 3 electrons (3 e⁻) to the left side gives a net charge of 0 on both sides: NO₃⁻ + 4 H⁺ + 3 e⁻ → NO + 2 H₂O

Electrons are added to the side with the higher net positive charge to satisfy conservation of charge.

Key Concept

Half-Reaction Method for Balancing Redox Equations in Acidic Medium

Estimated Time:1m 30s

Question 6Question →

Consider the redox reaction taking place in an acidic medium:

\text{MnO}_4^-(\text{aq}) + \text{H}_2\text{O}_2(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + \text{O}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest whole-number coefficients, what is the stoichiometric coefficient of

\text{H}^+(\text{aq})

?

16

6

8

3

Show answer & explanation

Answer: 6

Answer

6

The balanced net redox equation is

2\text{MnO}_4^-(\text{aq}) + 5\text{H}_2\text{O}_2(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{O}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})

. Thus, the stoichiometric coefficient of

\text{H}^+(\text{aq})

is 6.

Step-by-Step Solution

1

Write and balance the reduction half-reaction

\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Manganese is reduced from oxidation state +7 in

\text{MnO}_4^-

to +2 in

\text{Mn}^{2+}

, requiring 5 electrons. Charge and mass are balanced using

\text{H}^+

and

\text{H}_2\text{O}

.

2

Write and balance the oxidation half-reaction

\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-

Oxygen in

\text{H}_2\text{O}_2

is oxidized from -1 to 0 in

\text{O}_2

, releasing 2 electrons per molecule.

3

Equalize electron loss and gain

Multiply reduction half-reaction by 2 and oxidation half-reaction by 5:

2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10\text{e}^-

The least common multiple of 5 and 2 electrons transferred is 10.

4

Combine the half-reactions and simplify redundant species

2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}

Subtracting

10\text{H}^+

and

10\text{e}^-

from both sides yields a net coefficient of 6 for

\text{H}^+

on the reactant side.

Key Concept

Ion-Electron Method for Balancing Redox Equations in Acidic Medium

Question 7Question →

In the reduction half-reaction

\text{Fe}^{3+}(\text{aq}) + n e^- \rightarrow \text{Fe}(\text{s})

what is the value of

n

required to balance the electrical charge?

1

2

3

6

Show answer & explanation

Answer: $3$

Answer

The value of

n

is

3

.

To reduce one

\text{Fe}^{3+}

ion with a

+3

charge to a neutral iron atom with a

0

charge, exactly

3

electrons (each carrying a charge of

-1

) must be added to the reactant side so that the net charge on both sides equals

0

.

Step-by-Step Solution

1

Determine the charge on the reactant side and product side

The left side has one

\text{Fe}^{3+}

ion with a charge of

+3

. The right side has neutral

\text{Fe}(\text{s})

with a charge of

0

.

Charge conservation must be satisfied in a balanced half-reaction.

2

Calculate the number of negative electrons (

e^-

) required to balance net charge

+3 + n(-1) = 0 \implies n = 3

.

Each electron carries a single negative charge (

-1

).

Key Concept

Charge conservation in half-reactions

Question 8Question →

In hot, concentrated alkaline solutions, chlorine gas undergoes a disproportionation redox reaction according to the unbalanced equation:

\text{Cl}_2(\text{g}) + \text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + \text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})

When this equation is balanced using the smallest set of whole-number coefficients, what is the stoichiometric coefficient of hydroxide ions (

\text{OH}^-

) and the total number of moles of electrons transferred in the balanced equation?

6 hydroxide ions and 5 moles of electrons

6 hydroxide ions and 6 moles of electrons

12 hydroxide ions and 10 moles of electrons

3 hydroxide ions and 5 moles of electrons

Show answer & explanation

Answer: 6 hydroxide ions and 5 moles of electrons

Answer

The balanced equation requires 6 hydroxide ions and involves the transfer of 5 moles of electrons.

In the balanced redox equation

3\text{Cl}_2(\text{g}) + 6\text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + 5\text{Cl}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

, the stoichiometric coefficient of hydroxide ions is 6, and 5 moles of electrons are transferred per mole of reaction as written.

Step-by-Step Solution

1

Assign oxidation numbers to determine the oxidation and reduction half-reactions.

Elemental chlorine

\text{Cl}_2

has an oxidation number of

0

. In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

(oxidation). In

\text{Cl}^-

, chlorine has an oxidation state of

-1

(reduction).

Disproportionation involves the simultaneous oxidation and reduction of the same element.

2

Write and balance the oxidation half-reaction in basic medium.

\frac{1}{2}\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 5e^-

One chlorine atom increases in oxidation state from

0

to

+5

, releasing

5e^-

. Six

\text{OH}^-

ions balance the charge and oxygen/hydrogen mass.

3

Write and balance the reduction half-reaction.

\frac{1}{2}\text{Cl}_2 + e^- \rightarrow \text{Cl}^-

One chlorine atom decreases in oxidation state from

0

to

-1

, accepting

1e^-

.

4

Equalize electron transfer between half-reactions and combine.

Multiply the reduction half-reaction by 5:

\frac{5}{2}\text{Cl}_2 + 5e^- \rightarrow 5\text{Cl}^-

. Combine with the oxidation half-reaction:

3\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 5\text{Cl}^- + 3\text{H}_2\text{O}

. Total electrons transferred

n_e = 5

.

The number of electrons lost in oxidation must equal the number gained in reduction.

Key Concept

Balancing Disproportionation Redox Reactions in Basic Medium

Question 9Question →

Consider the unbalanced redox reaction taking place in an acidic medium:

\text{a MnO}_4^-(\text{aq}) + \text{b SO}_3^{2-}(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d Mn}^{2+}(\text{aq}) + \text{e SO}_4^{2-}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest set of whole-number coefficients, what is the value of the stoichiometric coefficient

c

for

\text{H}^+(\text{aq})

?

Show answer & explanation

Answer: 6

Answer

The value of the stoichiometric coefficient c for H+(aq) is 6.

Balancing the reduction half-reaction (

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

) and oxidation half-reaction (

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

) gives a combined total of

16\text{H}^+

on the reactant side and

10\text{H}^+

on the product side. Subtracting

10\text{H}^+

from both sides leaves a net coefficient of 6 for

\text{H}^+(\text{aq})

on the reactant side.

Step-by-Step Solution

1

Write the balanced reduction half-reaction for permanganate ion in acidic medium.

\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})

Manganese goes from oxidation state +7 to +2, requiring 5 electrons, 8 H+ ions to balance oxygen atoms, forming 4 H2O molecules.

2

Write the balanced oxidation half-reaction for sulfite ion to sulfate ion.

\text{SO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{SO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2e^-

Sulfur goes from oxidation state +4 to +6, releasing 2 electrons and 2 H+ ions while consuming 1 H2O molecule.

3

Equalize the number of transferred electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5.

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

The least common multiple of 5 and 2 transferred electrons is 10.

4

Combine the half-reactions and subtract common species (

10e^-

,

10\text{H}^+

, and

5\text{H}_2\text{O}

) from both sides.

2\text{MnO}_4^-(\text{aq}) + 5\text{SO}_3^{2-}(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{SO}_4^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

Subtracting

10\text{H}^+

from

16\text{H}^+

leaves

6\text{H}^+

on the reactant side, giving

c = 6

.

Key Concept

Balancing Redox Equations using the Ion-Electron Method in Acidic Medium

Question 10Question →

Consider the half-reaction representing the oxidation of thiosulfate ions to tetrathionate ions:

2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + n e^-

What is the number of electrons,

n

, required to balance the charge in this half-reaction?

Show answer & explanation

Answer: 2

Answer

The number of electrons required to balance the charge in the half-reaction is 2.

To balance a half-reaction, both atom counts and net electric charges must be equal on both sides of the equation. The reactant side contains 2 thiosulfate ions (

2\text{S}_2\text{O}_3^{2-}

), giving a net charge of

2 \times (-2) = -4

. The product side contains 1 tetrathionate ion (

\text{S}_4\text{O}_6^{2-}

), giving a net charge of

-2

. Adding 2 electrons (

2 e^-

) to the product side lowers its total charge to

-4

, equalizing the charge on both sides.

Step-by-Step Solution

1

Calculate the total charge of the reactant species.

Reactant charge = 2 * (-2) = -4.

There are 2 thiosulfate ions, each with an ionic charge of -2.

2

Calculate the net charge of the ionic product species.

Product charge (excluding electrons) = -2.

There is 1 tetrathionate ion with an ionic charge of -2.

3

Equate the overall charges on both sides to solve for the number of electrons n.

-4 = -2 - n, giving n = 2.

Adding 2 electrons (each carrying a -1 charge) to the product side brings the total product charge to -4, matching the reactant side.

Key Concept

Balancing electric charge in oxidation half-reactions

Question 11Question →

Consider the unbalanced redox reaction occurring in acidic solution:

\text{a AsO}_3^{3-}(\text{aq}) + \text{b MnO}_4^-(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d AsO}_4^{3-}(\text{aq}) + \text{e Mn}^{2+}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this ionic equation is balanced using the smallest possible whole-number coefficients, what is the value of the coefficient

c

for

\text{H}^+

?

Show answer & explanation

Answer: 6

Answer

The coefficient c for hydrogen ions (H⁺) in the balanced redox equation is 6.

Balancing the oxidation half-reaction shows that each arsenite ion produces 2 electrons and 2 H⁺ ions. The reduction half-reaction shows that each permanganate ion consumes 5 electrons and 8 H⁺ ions. Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 balances the total electron transfer at 10 electrons. Combining the equations gives 16 H⁺ on the left and 10 H⁺ on the right, which simplifies to 6 H⁺ on the reactant side.

Step-by-Step Solution

1

Balance the oxidation half-reaction (arsenite to arsenate)

AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻

Arsenic changes oxidation state from +3 to +5, releasing 2 electrons. Oxygen is balanced with H₂O and hydrogen with H⁺.

2

Balance the reduction half-reaction (permanganate to manganese(II))

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Manganese changes oxidation state from +7 to +2, consuming 5 electrons in acidic medium.

3

Equalize the electrons transferred in both half-reactions

5(AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻) and 2(MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)

The total number of electrons gained and lost must equal 10 electrons.

4

Combine the half-reactions and cancel common terms

5 AsO₃³⁻ + 2 MnO₄⁻ + 6 H⁺ → 5 AsO₄³⁻ + 2 Mn²⁺ + 3 H₂O

Subtracting 10 H⁺ and 5 H₂O from both sides leaves 6 H⁺ on the reactant side.

Key Concept

Balancing Ion-Electron Redox Half-Reactions in Acidic Medium

Balancing Redox Equations and Half-Reactions

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