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Chapter 10: Problem 43

Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{PCl}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) (c) \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{HNO}_{3}\) (e) \(\mathrm{HCOOH}\) (f) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)

Short Answer

Expert verified
The oxidation states of the given compounds are: (a) PCl$_3$: P: +3, Cl: -1 (b) H$_2$S: H: +1, S: -2 (c) MnO$_4^{-}$: Mn: +7, O: -2 (d) HNO$_3$: H: +1, N: +5, O: -2 (e) HCOOH: H: +1, C: 0/+2, O: -2 (f) S$_2$O$_3^{2-}$: S: +2/+4, O: -2

Step by step solution

01

(a) PCl3

For PCl3 (phosphorus trichloride), let's assign the oxidation state of P as x and for Cl as -1 (as per the rules for halogens). Using rule 3: x + 3(-1)= 0, solving for x we get, x = +3. So, the oxidation states are P: +3, Cl: -1.
02

(b) H2S

For H2S (hydrogen sulfide), let's assign the oxidation state of H as +1 (as per the rules for hydrogen) and for S as x. Using rule 3: 2(+1) + x = 0, solving for x we get, x = -2. So, the oxidation states are H: +1, S: -2.
03

(c) MnO4-

For MnO4- (permanganate ion), let's assign the oxidation state of Mn as x and for O as -2 (as per the rules for oxygen). Using rule 3: x + 4(-2) = -1, solving for x we get, x = +7. So, the oxidation states are Mn: +7, O: -2.
04

(d) HNO3

For HNO3 (nitric acid), let's assign the oxidation state of H as +1 (as per the rules for hydrogen), for N as x and for O as -2 (as per the rules for oxygen). Using rule 3: +1 + x + 3(-2) = 0, solving for x we get, x = +5. So, the oxidation states are H: +1, N: +5, O: -2.
05

(e) HCOOH

For HCOOH (formic acid), let's assign the oxidation state of H as +1 (as per the rules for hydrogen), for C in both instances as x, and for O as -2 (as per the rules for oxygen). Using rule 3: +1 + x + 2(-2) + x + 1 = 0, solving for x we get, x = 0 (for the C bonded to the H), x = +2 (for the C bonded to the O). So, the oxidation states are H: +1, C: 0/+2, O: -2.
06

(f) S2O3^2-

For S2O3^2- (thiosulfate ion), let's assign the oxidation state of S as x (for both sulfur atoms) and for O as -2 (as per the rules for oxygen). Using rule 3: 2x + 3(-2) = -2, solving for x we get, x = +2 (for the S bonded to the other S atom), x = +4 (for the S bonded to the three O atoms). So, the oxidation states are S: +2/+4, O: -2.

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Most popular questions from this chapter

If you put a piece of iron in an aqueous solution of blue \(\mathrm{Cu}^{2+}\) ions, the spontaneous redox reaction \(\mathrm{Fe}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Fe}^{2+}+\mathrm{Cu}\) will occur. An aqueous solution of \(\mathrm{Fe}^{2+}\) ions is red-brown. (a) What is oxidized? (b) What is reduced? (c) What is the oxidizing agent? (d) What is the reducing agent? (e) What will happen to the piece of iron over time? (f) Do the electrons move from the oxidizing agent to the reducing agent or from the reducing agent to the oxidizing agent? (g) How could you use this reaction to make a battery? (Explain and show your battery in a diagram, using a nail and a penny as your source of iron and copper, respectively.)

When you turn on an electrical appliance, are you consuming electrons?

Use the shortcut rules to assign oxidation states to all atoms. \(\mathrm{CO}_{3}^{2-}\)

Which is a better reducing agent, \(\mathrm{Zn}\) or \(\mathrm{Pb}\) ?

A battery is constructed from tin and copper by dipping strips of each metal into a solution of its ions \(\left(\mathrm{Sn}^{2+}\right.\) and \(\mathrm{Cu}^{2+}\), respectively). As the battery operates, the \(\mathrm{Sn}^{2+}\) concentration increases and the \(\mathrm{Cu}^{2+}\) concentration decreases. (a) What is getting oxidized? (b) What is getting reduced? (c) Draw a battery similar to the one you drew for WorkPatch 10.7. Make sure you label which way the electrons flow, and also label the cathode, \(+\), and anode, \(=\)
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