Electrochemistry

Sample questions

Showing 10 of 70 questions from this chapter. The full bank is in the chapter test below.

1. 1

   Which process is defined as the loss of one or more electrons by a species?

   • A.✓ Oxidation (correct answer)
   • B.Reduction
   • C.Disproportionation
   • D.Neutralization

   Why: Oxidation is the loss of electrons, as in $\ce{Zn -> Zn2+ + 2e-}$. Reduction is the opposite process, the gain of electrons.
2. 2

   The gain of one or more electrons by a species is called:

   • A.Oxidation
   • B.Disproportionation
   • C.✓ Reduction (correct answer)
   • D.Electrolysis

   Why: Reduction is the gain of electrons, as in $\ce{Cu2+ + 2e- -> Cu}$. A species that gains electrons has its oxidation number lowered.
3. 3

   In the change $\ce{Fe2+ -> Fe3+ + e-}$, the $\ce{Fe2+}$ ion is:

   • A.Reduced, because it gains an electron
   • B.✓ Oxidized, because it loses an electron (correct answer)
   • C.Unchanged, because its charge hardly alters
   • D.Reduced, because its oxidation number falls

   Why: $\ce{Fe2+}$ loses an electron, so it is oxidized and its oxidation number rises from +2 to +3. The loss of an electron can never be reduction.
4. 4

   In the half-equation $\ce{Cl2(g) + 2e- -> 2Cl-(aq)}$, chlorine is:

   • A.Oxidized from 0 to -1
   • B.Acting as a reducing agent
   • C.Unchanged in oxidation number
   • D.✓ Reduced from 0 to -1 (correct answer)

   Why: Each chlorine atom gains an electron, so its oxidation number falls from 0 to -1 and chlorine is reduced. A species that gains electrons is an oxidizing agent, not a reducing agent.
5. 5

   Photosynthesis in green plants and respiration in living cells are both examples of:

   • A.Precipitation reactions
   • B.Acid-base reactions
   • C.✓ Redox reactions (correct answer)
   • D.Neutralization reactions

   Why: Both processes involve the transfer of electrons between species, so both are redox reactions. Photosynthesis stores energy while respiration releases it, but in each case electron transfer is taking place.
6. 6

   An oxidation number is best described as:

   • A.The actual number of electrons present in the outer shell of an atom
   • B.✓ The apparent charge on an atom of an element in a molecule or an ion (correct answer)
   • C.The number of protons present in the nucleus of a single atom
   • D.The number of covalent bonds an atom forms in a compound

   Why: An oxidation number is the apparent charge on one atom of an element in a molecule or ion, and it may be positive, negative or zero. It is not the true electron count and not the number of bonds.
7. 7

   The oxidation number of each sulfur atom in $\ce{S8}$ is:

   • A.-2
   • B.+6
   • C.+2
   • D.✓ 0 (correct answer)

   Why: The oxidation number of any uncombined element is zero, so every atom in $\ce{S8}$, $\ce{H2}$ or $\ce{Cl2}$ has an oxidation number of 0. The value -2 belongs to sulfide ions in compounds such as $\ce{Na2S}$.
8. 8

   Oxygen has an oxidation number of -1/2 in:

   • A.✓ $\ce{KO2}$ (correct answer)
   • B.$\ce{H2O}$
   • C.$\ce{Na2O2}$
   • D.$\ce{F2O}$

   Why: Oxygen is -1/2 in $\ce{KO2}$, -2 in $\ce{H2O}$, -1 in $\ce{Na2O2}$ and +2 in $\ce{F2O}$. The positive value in $\ce{F2O}$ arises because fluorine is more electronegative than oxygen.
9. 9

   The oxidation number of hydrogen in sodium hydride, NaH, is:

   • A.+1
   • B.✓ -1 (correct answer)
   • C.0
   • D.+2

   Why: In metal hydrides such as NaH the more electronegative element is hydrogen, so hydrogen takes the negative oxidation number -1. Hydrogen is +1 in its compounds with non-metals.
10. 10

    In $\ce{F2O}$, the oxidation number of oxygen is:

    • A.-2
    • B.-1
    • C.✓ +2 (correct answer)
    • D.0

    Why: Fluorine is more electronegative than oxygen, so oxygen is the positive partner here and has an oxidation number of +2. The usual -2 value applies only when oxygen is the more electronegative element in the compound.