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In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g)-\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow\) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(\begin{aligned} 3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) & \\ & 3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}\) (d) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q)--\rightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
In each oxidation-reduction equation: a) Iodine (I) decreases its oxidation number by 5 and Carbon (C) increases its oxidation number by 2. b) Mercury (Hg) decreases its oxidation number by 2 and Nitrogen (N) increases its oxidation number by 2. c) Sulfur (S) increases its oxidation number by 2 and Nitrogen (N) decreases its oxidation number by 3. d) Oxygen (O) in H鈧侽鈧 increases its oxidation number by 1, while the other elements' oxidation numbers remain unchanged.

Step by step solution

01

a) I鈧侽鈧(s) + 5 CO(g) 鉄 I鈧(s) + 5 CO鈧(g)

Determine the oxidation numbers of the elements: - I鈧侽鈧: Iodine(I) = +5, Oxygen(O) = -2 - CO: Carbon(C) = +2, Oxygen(O) = -2 - I鈧: Iodine(I) = 0 - CO鈧: Carbon(C) = +4, Oxygen(O) = -2 Now compare the changes in oxidation numbers: - Iodine(I): From +5 to 0, a decrease of 5. - Carbon(C): From +2 to +4, an increase of 2. The other elements' oxidation numbers remain unchanged.
02

b) 2 Hg^(2+)(aq) + N鈧侶鈧(aq) 鉄 2 Hg(l) + N鈧(g) + 4 H^(+)(aq)

Determine the oxidation numbers of the elements: - Hg^(2+): Mercury(Hg) = +2 - N鈧侶鈧: Nitrogen(N) = -2, Hydrogen(H) = +1 - Hg: Mercury(Hg) = 0 - N鈧: Nitrogen(N) = 0 - H^(+): Hydrogen(H) = +1 Now compare the changes in oxidation numbers: - Mercury(Hg): From +2 to 0, a decrease of 2. - Nitrogen(N): From -2 to 0, an increase of 2. - Hydrogen(H) remains unchanged.
03

c) 3 H鈧係(aq) + 2 H^(+)(aq) + 2 NO鈧僞(-)(aq) 鉄 3 S(s) + 2 NO(g) + 4 H鈧侽(l)

Determine the oxidation numbers of the elements: - H鈧係: Hydrogen(H) = +1, Sulfur(S) = -2 - H^(+): Hydrogen(H) = +1 - NO鈧僞(-): Nitrogen(N) = +5, Oxygen(O) = -2 - S: Sulfur(S) = 0 - NO: Nitrogen(N) = +2, Oxygen(O) = -2 - H鈧侽: Hydrogen(H) = +1, Oxygen(O) = -2 Now compare the changes in oxidation numbers: - Sulfur(S): From -2 to 0, an increase of 2. - Nitrogen(N): From +5 to +2, a decrease of 3. - The other elements' oxidation numbers remain unchanged.
04

d) Ba^(2+)(aq) + 2 OH^(-)(aq) + H鈧侽鈧(aq) + 2 ClO鈧(aq) 鉄 Ba(ClO鈧)鈧(s) + 2 H鈧侽(l) + O鈧(g)

Determine the oxidation numbers of the elements: - Ba^(2+): Barium(Ba) = +2 - OH^(-): Oxygen(O) = -2, Hydrogen(H) = +1 - H鈧侽鈧: Oxygen(O) = -1, Hydrogen(H) = +1 - ClO鈧: Chlorine(Cl) = +3, Oxygen(O) = -2 - Ba(ClO鈧)鈧: Barium(Ba) = +2, Chlorine(Cl) = +3, Oxygen(O) = -2 - H鈧侽: Oxygen(O) = -2, Hydrogen(H) = +1 - O鈧: Oxygen(O) = 0 Now compare the changes in oxidation numbers: - Oxygen(O) in H鈧侽鈧: From -1 to 0, an increase of 1. - Chlorine(Cl): From +3 to +3, no change. - The other elements' oxidation numbers remain unchanged.

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Key Concepts

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

Changes in Oxidation Number
Understanding changes in oxidation number is crucial when studying oxidation-reduction (redox) reactions. The oxidation number, often referred to as oxidation state, can be considered the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic.

An increase in oxidation number signifies oxidation and typically involves the loss of electrons. Conversely, a decrease in oxidation number indicates reduction, corresponding with a gain of electrons. Detecting these changes allows for the identification of which elements are oxidized and which are reduced in a redox reaction.

For example, in the reaction between I鈧侽鈧 and CO to form I鈧 and CO鈧, iodine is reduced (oxidation number decreases from +5 to 0), and carbon is oxidized (oxidation number increases from +2 to +4). By tracking these changes, we get insights into the electron transfer processes underlying redox reactions.
Balancing Redox Equations
Balancing redox equations involves ensuring that both mass and charge are conserved during the reaction. This can be more complex than balancing standard chemical equations, as it requires balancing not just the number of atoms but also the electrons lost in oxidation and gained in reduction.

To balance a redox equation, one can use the half-reaction method, which separately balances the reduction and oxidation reactions before combining them. It's important to adjust coefficients to balance the number of electrons transferred and to ensure the same number of electrons are involved in both half-reactions.

Tips for Balancing Redox Equations

  • Assign oxidation states to determine which species are oxidized and reduced.
  • Separate the reaction into half-reactions for oxidation and reduction.
  • Balance each half-reaction for mass and charge, adding H鈧侽, H鈦, and electrons as necessary.
  • Combine the half-reactions, making sure the number of electrons cancels out.
  • Verify that both mass and charge are balanced in the final equation.
Properly balancing redox equations is imperative not just for academic exercises, but also for practical applications in chemistry and industry.
Oxidation States
Oxidation states are a fundamental concept in the study of chemistry, particularly in redox reactions. An element's oxidation state is a numerical indicator of its degree of oxidation or reduction; it essentially represents the number of electrons an atom can gain, lose, or share when it forms chemical compounds.

Determining Oxidation States:

  • The oxidation state of an atom in its elemental form is always zero.
  • For a simple (monoatomic) ion, the oxidation state is equal to the charge of the ion.
  • Oxygen generally has an oxidation state of -2, except in peroxides where it is -1, and in compounds bonded to fluorine, where it can be positive.
  • Hydrogen generally has an oxidation state of +1, except when bonded to metals in hydrides where it is -1.
  • The sum of the oxidation states for all atoms in a neutral molecule or formula unit equals zero; for an ion, it equals the charge of the ion.
Mastering the assignment of oxidation states is essential, as it enables students to understand electron transfer reactions and to predict the outcomes of chemical reactions.

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Most popular questions from this chapter

(a) Write the reactions for the discharge and charge of a nickel-cadmium rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\). The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode, or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g)-\rightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=2.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.85 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both compartments is \(5.00 ?\)

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules? (c) What is special about a standard cell potential?

(a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metalhydride batteries over the nickel-cadmium batteries?

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