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Chapter 16: Problem 566

Balance the following reaction in basic aqueous solution: \(\mathrm{SO}^{2-}{ }_{3}+\mathrm{CrO}^{2-}{ }_{4} \rightarrow \mathrm{SO}^{2-}{ }_{4}+\mathrm{Cr}(\mathrm{OH})_{3}\)

Short Answer

Expert verified
The balanced chemical equation in basic aqueous solution is: \[ 3\mathrm{SO}_{3}^{2-} + \mathrm{CrO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 3\mathrm{SO}_{4}^{2-} + \mathrm{Cr}(\mathrm{OH})_{3} + 14\mathrm{OH}^- \]

Step by step solution

01

Identify the Oxidation States

First, determine the oxidation states of all elements in the reactants and products. Sulfur (\(S\)): In \(\mathrm{SO}^{2-}{ }_{3}\), sulfur has an oxidation state of +4. In \(\mathrm{SO}^{2-}{ }_{4}\), sulfur has an oxidation state of +6. Chromium (\(Cr\)): In \(\mathrm{CrO}^{2-}{ }_{4}\), chromium has an oxidation state of +6. In \(\mathrm{Cr}(\mathrm{OH})_{3}\), chromium has an oxidation state of +3.
02

Write the Two Half-Reactions

Using the identification of oxidation states, we can write down the two half-reactions for the redox process. Oxidation (Sulfur): \[ \mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-} \] Reduction (Chromium): \[ \mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Cr}(\mathrm{OH})_{3} \]
03

Balance Atoms and Charges in Half-Reactions

Apart from oxygen and hydrogen, balance the atoms for each half-reaction: Oxidation half-reaction: \[ \mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-} \] The sulfur atoms are already balanced. Reduction half-reaction: \[ \mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Cr}(\mathrm{OH})_{3} \] The chromium atoms are already balanced. Next, we balance the oxygens using \(\mathrm{H}_{2}\mathrm{O}\): Oxidation half-reaction: \[ \mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \] Reduction half-reaction: \[ \mathrm{CrO}_{4}^{2-} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Cr}(\mathrm{OH})_{3} \] Then, balance the hydrogen and charge using \(\mathrm{OH}^-\) and electrons (\(e^-\)): Oxidation half-reaction: \[ \mathrm{SO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{SO}_{4}^{2-} + 2\mathrm{H}^+ + 2e^- \] Reduction half-reaction: \[ \mathrm{CrO}_{4}^{2-} + 4\mathrm{H}_{2}\mathrm{O} + 6e^- \rightarrow \mathrm{Cr}(\mathrm{OH})_{3} + 8\mathrm{OH}^- \]
04

Combine the Half-Reactions

To combine the half-reactions, we must equalize the number of electrons in both half-reactions and then add them up. In this case, multiply the oxidation half-reaction by 3 so that the numbers of electrons are equal: 3(Oxidation half-reaction): \[ 3\mathrm{SO}_{3}^{2-} + 3\mathrm{H}_{2}\mathrm{O} \rightarrow 3\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^+ + 6e^- \] Adding up the half-reactions: \[ 3\mathrm{SO}_{3}^{2-} + 3\mathrm{H}_{2}\mathrm{O} + \mathrm{CrO}_{4}^{2-} + 4\mathrm{H}_{2}\mathrm{O} + 6e^- \rightarrow 3\mathrm{SO}_{4}^{2-} + 6\mathrm{H}^+ + 6e^- + \mathrm{Cr}(\mathrm{OH})_{3} + 8\mathrm{OH}^- \]
05

Simplify the Reaction

Simplify the reaction by canceling out the common components and combining the like terms: \[ 3\mathrm{SO}_{3}^{2-} + \mathrm{CrO}_{4}^{2-} + 7\mathrm{H}_{2}\mathrm{O} \rightarrow 3\mathrm{SO}_{4}^{2-} + \mathrm{Cr}(\mathrm{OH})_{3} + 6\mathrm{H}^+ + 8\mathrm{OH}^- \] Now, cancel out the \(\mathrm{H}^+\) and \(\mathrm{OH}^-\) using \(6\mathrm{H}_{2}\mathrm{O}\): \[ 3\mathrm{SO}_{3}^{2-} + \mathrm{CrO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 3\mathrm{SO}_{4}^{2-} + \mathrm{Cr}(\mathrm{OH})_{3} + 14\mathrm{OH}^- \] And you'll get the balanced chemical equation in basic aqueous solution: \[ 3\mathrm{SO}_{3}^{2-} + \mathrm{CrO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 3\mathrm{SO}_{4}^{2-} + \mathrm{Cr}(\mathrm{OH})_{3} + 14\mathrm{OH}^- \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
Understanding oxidation states is key to identifying changes in a redox reaction. Oxidation states help us track the transfer of electrons. To find the oxidation state, we assign each atom a number that reflects its electron count compared to its neutral state.

Here鈥檚 a quick guide on determining oxidation states:
  • Elements in their natural form have an oxidation state of 0.
  • For ions, the oxidation state is equal to the charge of the ion.
  • Oxygen usually has an oxidation state of -2, and hydrogen is +1 in compounds.
In our example, sulfur goes from +4 in \( \mathrm{SO}^{2-}{ }_{3} \) to +6 in \( \mathrm{SO}^{2-}{ }_{4} \), indicating oxidation, while chromium goes from +6 in \( \mathrm{CrO}^{2-}{ }_{4} \) to +3 in \( \mathrm{Cr(OH)}_{3} \), showing reduction. Recognizing these changes helps us set up the half-reactions.
Half-Reactions
Half-reactions separate the oxidation and reduction processes in a redox reaction. They allow us to balance each part separately, making it easier to handle complex equations.

In the given exercise, the two half-reactions are:
  • **Oxidation of Sulfur:** \(\mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-} \)
  • **Reduction of Chromium:** \(\mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Cr(OH)}_{3} \)
Each half-reaction is balanced by:
  • Balancing atoms other than O and H first.
  • Balancing oxygen using \( \mathrm{H}_{2}\mathrm{O} \).
  • Balancing hydrogen using \( \mathrm{H}^{+} \) for acidic or \( \mathrm{OH}^{-} \) for basic solutions.
  • Balancing charge using electrons (\( e^- \)).
After balancing, these half-reactions are combined, ensuring the electron loss equals the electron gain.
Basic Aqueous Solution
A basic aqueous solution provides a specific context for balancing redox equations. These solutions contain an excess of \( \mathrm{OH}^- \) ions, which means special steps must be taken to balance equations properly.

For reactions occurring in basic conditions, once the half-reactions are balanced in acidic conditions, we add \( \mathrm{OH}^- \) to both sides to neutralize any \( \mathrm{H}^{+} \), forming water. This might lead to the cancellation of water molecules on both sides.

The reaction from our exercise simplifies as follows:
  • Add \( \mathrm{OH}^- \) to neutralize \( \mathrm{H}^{+} \) after combining the half-reactions.
  • Combine and cancel water if it appears on both sides.
This results in a perfectly balanced equation under basic conditions.
Chemical Equation Balancing
Balancing a chemical equation ensures conservation of mass and charge. Each atom and charge must be equal on both sides of the equation.

Steps to achieve a balanced equation include:
  • Determining oxidation states to identify the redox pairs.
  • Setting up separate oxidation and reduction half-reactions.
  • Balancing atoms and charges as detailed in half-reaction balancing.
  • Combining the half-reactions, ensuring equal electron transfer.
  • For basic solutions, adjust with \( \mathrm{OH}^- \) and simplify.
The exercise solution demonstrates these steps by combining and simplifying half-reactions, resulting in a fully balanced redox equation. This principle applies universally, aiding in understanding complex chemical reactions.

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Most popular questions from this chapter

Balance the equation for the following reaction taking place in aqueous basic solution: \(\mathrm{MnO}^{-}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{MnO}_{2}+\mathrm{O}_{2}\)

Given the following standard electrode potentials at \(25^{\circ} \mathrm{C}\) : \(\mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+}, \mathrm{E}^{\circ}=0.15 \mathrm{ev}\) and \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\) \(E^{\circ}=+0.77 e \mathrm{ev}\) will the reaction \(\mathrm{Sn}^{2+}+2 \mathrm{~F} \mathrm{e}^{3+} \leftrightharpoons \mathrm{Sn}^{4+}+2 \mathrm{Fe}^{2+}\) proceed spontaneously?

Determine the volume in milliliters of \(20 \mathrm{M} \mathrm{KMnO}_{4}\) required to oxidize \(25.0 \mathrm{ml}\) of \(40 \mathrm{M} \mathrm{FeSO}_{4}\), in acidic solution. Assume the reaction which occurs is the oxidation of \(\mathrm{Fe}^{2+}\) by \(\mathrm{MnO}^{-}{ }_{4}\) to give \(\mathrm{Fe}^{+3}\) and \(\mathrm{Mn}^{2+}\).

For the following voltaic cell, write the half-reactions, designating which is oxidation and which is reduction. Write the cell reaction and calculate the voltage of the cell made from standard electrodes. The cell is \(\mathrm{Co} ; \mathrm{Co}^{+2} \mid \mathrm{Ni}^{+2} ; \mathrm{Ni}\).

Two electrolytic cells were placed in series. One was composed of \(\mathrm{AgNO}_{3}\) and the other of \(\mathrm{CuSO}_{4}\). Electricity was passed through the cells until \(1.273 \mathrm{~g}\) of \(\mathrm{Ag}\) had been deposited. How much copper was deposited at the same time?
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