Specify which of the following equations represent oxidationreduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\) c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
In summary: a. Redox reaction - Oxidizing agent: \(\mathrm{H}_{2}\mathrm{O}\) - Reducing agent: \(\mathrm{CH}_{4}\) - Species oxidized: C in \(\mathrm{CH}_{4}\) - Species reduced: O in \(\mathrm{H}_{2} \mathrm{O}\) b. Redox reaction - Oxidizing agent: \(\mathrm{AgNO}_{3}\) - Reducing agent: \(\mathrm{Cu}\) - Species oxidized: Cu - Species reduced: Ag c. Redox reaction - Oxidizing agent: \(\mathrm{HCl}\) - Reducing agent: \(\mathrm{Zn}\) - Species oxidized: Zn - Species reduced: H d. Not a redox reaction

Step by step solution

01

Determine the oxidation states of each element

For reactants: C(-4), H(+1), O(-2) For products: C(+2), H(+1), O(-2)
02

Check for changes in oxidation states

Yes, there are changes in oxidation states: C from -4 to +2 (oxidation) O from -2 to -2 (no change) H from +1 to +1 (no change)
03

Identify the redox agents and species

Oxidizing agent: \(\mathrm{H}_{2}\mathrm{O}\) (contains species reduced, O) Reducing agent: \(\mathrm{CH}_{4}\) (contains species oxidized, C) Species oxidized: C in \(\mathrm{CH}_{4}\) Species reduced: O in \(\mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)\)
04

Determine the oxidation states of each element

For reactants: Ag(+1), N(+5), O(-2), Cu(0) For products: Cu(+2), N(+5), O(-2), Ag(0)
05

Check for changes in oxidation states

Yes, there are changes in oxidation states: Ag from +1 to 0 (reduction) Cu from 0 to +2 (oxidation) N from +5 to +5 (no change) O from -2 to -2 (no change)
06

Identify the redox agents and species

Oxidizing agent: \(\mathrm{AgNO}_{3}\) (contains species reduced, Ag) Reducing agent: \(\mathrm{Cu}\) (contains species oxidized, Cu) Species oxidized: Cu Species reduced: Ag c. \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\)
07

Determine the oxidation states of each element

For reactants: Zn(0), H(+1), Cl(-1) For products: Zn(+2), Cl(-1), H(0)
08

Check for changes in oxidation states

Yes, there are changes in oxidation states: Zn from 0 to +2 (oxidation) H from +1 to 0 (reduction) Cl from -1 to -1 (no change)
09

Identify the redox agents and species

Oxidizing agent: \(\mathrm{HCl}\) (contains species reduced, H) Reducing agent: \(\mathrm{Zn}\) (contains species oxidized, Zn) Species oxidized: Zn Species reduced: H d. \(2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)
10

Determine the oxidation states of each element

For reactants: H(+1), Cr(+6), O(-2) For products: Cr(+6), O(-2), H(+1), O(-2)
11

Check for changes in oxidation states

No, there are no changes in oxidation states: H from +1 to +1 (no change) Cr from +6 to +6 (no change) O from -2 to -2 (no change) Since there are no changes in oxidation states in reaction (d), it is not a redox reaction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{~V} \end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{~kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

Calculate \(K_{\mathrm{sp}}\) for iron(II) sulfide given the following data: $$\begin{aligned}\mathrm{FeS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)+\mathrm{S}^{2-}(a q) & & \mathscr{b}^{\circ} &=-1.01 \mathrm{~V} \\\\\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & & \mathscr{b}^{\circ} &=-0.44 \mathrm{~V} \end{aligned}$$

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free