301. A table lists possible reduction products of \( \mathrm{MnO_4^-} \).
| Medium | Reduction product | \(n\)-factor of \( \mathrm{MnO_4^-} \) |
|---|
| P. Acidic | \( \mathrm{Mn^{2+}} \) | \(5\) |
| Q. Neutral or mildly basic | \( \mathrm{MnO_2} \) | \(3\) |
| R. Strongly basic | \( \mathrm{MnO_4^{2-}} \) | \(1\) |
| S. Acidic | \( \mathrm{Mn^{2+}} \) | \(3\) |
Which row needs correction?
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: In acidic medium, manganese changes from \(+7\) in \( \mathrm{MnO_4^-} \) to \(+2\) in \( \mathrm{Mn^{2+}} \), so \(n=5\). In neutral or mildly basic medium, the product \( \mathrm{MnO_2} \) has manganese \(+4\), giving \(n=3\). In strongly basic medium, \( \mathrm{MnO_4^{2-}} \) has manganese \(+6\), so \(n=1\). Row S repeats the acidic product but gives the neutral/basic \(n\)-factor. The medium and product must be read together.
302. For \( \mathrm{Cr_2O_7^{2-}\rightarrow 2Cr^{3+}} \) in acidic medium, the \(n\)-factor of \( \mathrm{Cr_2O_7^{2-}} \) is:
ⓐ. \(6\)
ⓑ. \(3\)
ⓒ. \(4\)
ⓓ. \(7\)
Correct Answer: \(6\)
Explanation: \( \textbf{Chromium in } \mathrm{Cr_2O_7^{2-}}: \)
\[
2x+7(-2)=-2
\]
\[
2x=+12
\]
\[
x=+6
\]
\( \textbf{Chromium in product:} \) \( \mathrm{Cr^{3+}} \) has oxidation number \(+3\).
\( \textbf{Change per chromium atom:} \)
\[
+6\rightarrow+3
\]
\( \textbf{Electron gain per chromium atom:} \) \(3\) electrons.
\( \textbf{Two chromium atoms:} \)
\[
2\times3=6
\]
\( \textbf{Final answer:} \) The \(n\)-factor of one \( \mathrm{Cr_2O_7^{2-}} \) ion is \(6\). Counting both chromium atoms is essential.
303. The equivalent mass of an oxidising or reducing agent in a given redox reaction is calculated by:
ⓐ. \( \frac{\text{molar mass}}{n\text{-factor}} \)
ⓑ. \( \text{molar mass}\times n\text{-factor} \)
ⓒ. \( \frac{n\text{-factor}}{\text{molar mass}} \)
ⓓ. \( \text{molar mass}+n\text{-factor} \)
Correct Answer: \( \frac{\text{molar mass}}{n\text{-factor}} \)
Explanation: Equivalent mass connects molar mass with the number of electrons exchanged per formula unit. In redox reactions, that electron count is the \(n\)-factor. The relation is \( \text{equivalent mass}=\frac{\text{molar mass}}{n} \). Multiplying by \(n\) would give a value too large, especially for oxidants such as \( \mathrm{KMnO_4} \) in acid where \(n=5\). The formula must be used only after the reaction-specific \(n\)-factor is known.
304. The molar mass of \( \mathrm{KMnO_4} \) is \(158\,\mathrm{g\,mol^{-1}}\). In acidic medium, \( \mathrm{MnO_4^-} \) is reduced to \( \mathrm{Mn^{2+}} \). The equivalent mass of \( \mathrm{KMnO_4} \) is:
ⓐ. \(52.7\,\mathrm{g\,equiv^{-1}}\)
ⓑ. \(31.6\,\mathrm{g\,equiv^{-1}}\)
ⓒ. \(79.0\,\mathrm{g\,equiv^{-1}}\)
ⓓ. \(158\,\mathrm{g\,equiv^{-1}}\)
Correct Answer: \(31.6\,\mathrm{g\,equiv^{-1}}\)
Explanation: \( \textbf{Given molar mass:} \)
\[
M(\mathrm{KMnO_4})=158\,\mathrm{g\,mol^{-1}}
\]
\( \textbf{Medium and product:} \) In acidic medium, \( \mathrm{MnO_4^-} \) changes to \( \mathrm{Mn^{2+}} \).
\( \textbf{Electron change:} \)
\[
\mathrm{Mn}:+7\rightarrow+2
\]
\( \textbf{\(n\)-factor:} \)
\[
n=5
\]
\( \textbf{Equivalent mass relation:} \)
\[
\text{equivalent mass}=\frac{\text{molar mass}}{n}
\]
\( \textbf{Substitution:} \)
\[
\text{equivalent mass}=\frac{158}{5}
\]
\[
=31.6\,\mathrm{g\,equiv^{-1}}
\]
\( \textbf{Final answer:} \) The equivalent mass is \(31.6\,\mathrm{g\,equiv^{-1}}\). Using \(158\,\mathrm{g\,equiv^{-1}}\) would ignore the five-electron reduction in acid.
305. The molar mass of \( \mathrm{K_2Cr_2O_7} \) is \(294\,\mathrm{g\,mol^{-1}}\). In acidic medium, \( \mathrm{Cr_2O_7^{2-}} \) is reduced to \( \mathrm{Cr^{3+}} \). Its equivalent mass is:
ⓐ. \(29.4\,\mathrm{g\,equiv^{-1}}\)
ⓑ. \(98.0\,\mathrm{g\,equiv^{-1}}\)
ⓒ. \(49.0\,\mathrm{g\,equiv^{-1}}\)
ⓓ. \(147\,\mathrm{g\,equiv^{-1}}\)
Correct Answer: \(49.0\,\mathrm{g\,equiv^{-1}}\)
Explanation: \( \textbf{Molar mass:} \)
\[
M(\mathrm{K_2Cr_2O_7})=294\,\mathrm{g\,mol^{-1}}
\]
\( \textbf{Chromium change:} \)
\[
\mathrm{Cr}:+6\rightarrow+3
\]
\( \textbf{Change per chromium atom:} \) \(3\) electrons are gained.
\( \textbf{Two chromium atoms per formula unit:} \)
\[
n=2\times3=6
\]
\( \textbf{Equivalent mass:} \)
\[
\frac{294}{6}=49.0\,\mathrm{g\,equiv^{-1}}
\]
\( \textbf{Final answer:} \) The equivalent mass is \(49.0\,\mathrm{g\,equiv^{-1}}\). Dividing by \(3\) would forget that one dichromate ion contains two chromium atoms.
306. A claim says, “The \(n\)-factor of \( \mathrm{KMnO_4} \) is always \(5\).” The best correction is:
ⓐ. \(n\)-factor is always equal to the number of oxygen atoms
ⓑ. \(n\)-factor is always equal to the ionic charge of \( \mathrm{MnO_4^-} \)
ⓒ. \(n\)-factor is not used in redox calculations
ⓓ. \(n\)-factor depends on the reduction product and medium
Correct Answer: \(n\)-factor depends on the reduction product and medium
Explanation: \( \mathrm{KMnO_4} \) has \(n=5\) when \( \mathrm{MnO_4^-} \) is reduced to \( \mathrm{Mn^{2+}} \) in acidic medium. It has \(n=3\) when reduced to \( \mathrm{MnO_2} \) in neutral or mildly basic medium. It may have \(n=1\) when reduced to \( \mathrm{MnO_4^{2-}} \) in strongly basic medium. The same formula can therefore have different \(n\)-factors in different reactions. The product and medium must be known before assigning \(n\).
307. Normality \( \left(N\right) \) and molarity \( \left(M\right) \) are related in a redox reaction by:
ⓐ. \(N=M\times n\)
ⓑ. \(N=\frac{M}{n}\)
ⓒ. \(N=M+n\)
ⓓ. \(N=M-n\)
Correct Answer: \(N=M\times n\)
Explanation: Normality measures equivalents per litre, while molarity measures moles per litre. In a redox reaction, one mole of a species gives \(n\) equivalents if its \(n\)-factor is \(n\). Therefore \(N=M\times n\). This relation is useful only when the correct reaction-specific \(n\)-factor is used. A normality value without its redox reaction context can be misleading.
308. A \( \mathrm{0.0200\,mol\,L^{-1}} \) acidified \( \mathrm{KMnO_4} \) solution is used where \( \mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{2+}} \). What is its normality?
ⓐ. \(0.00400\,\mathrm{N}\)
ⓑ. \(0.0200\,\mathrm{N}\)
ⓒ. \(0.100\,\mathrm{N}\)
ⓓ. \(0.0600\,\mathrm{N}\)
Correct Answer: \(0.100\,\mathrm{N}\)
Explanation: \( \textbf{Given molarity:} \)
\[
M=0.0200\,\mathrm{mol\,L^{-1}}
\]
\( \textbf{Redox change in acid:} \)
\[
\mathrm{Mn}:+7\rightarrow+2
\]
\( \textbf{\(n\)-factor:} \)
\[
n=5
\]
\( \textbf{Normality relation:} \)
\[
N=M\times n
\]
\( \textbf{Substitution:} \)
\[
N=0.0200\times5
\]
\[
N=0.100\,\mathrm{N}
\]
\( \textbf{Final answer:} \) The normality is \(0.100\,\mathrm{N}\). The same molarity would give a different normality if permanganate formed a different reduction product.
309. Read the data record below.
A \( \mathrm{0.100\,N} \) oxidising-agent solution reacts with a reducing-agent solution. \( \mathrm{20.0\,mL} \) of the oxidising-agent solution is exactly equivalent to \( \mathrm{25.0\,mL} \) of the reducing-agent solution.
What is the normality of the reducing-agent solution?
ⓐ. \(0.100\,\mathrm{N}\)
ⓑ. \(0.0800\,\mathrm{N}\)
ⓒ. \(0.125\,\mathrm{N}\)
ⓓ. \(0.200\,\mathrm{N}\)
Correct Answer: \(0.0800\,\mathrm{N}\)
Explanation: \( \textbf{Equivalent relation at equivalence:} \)
\[
N_1V_1=N_2V_2
\]
\( \textbf{Known values:} \)
\[
N_1=0.100\,\mathrm{N}
\]
\[
V_1=20.0\,\mathrm{mL}
\]
\[
V_2=25.0\,\mathrm{mL}
\]
\( \textbf{Solve for } N_2: \)
\[
N_2=\frac{N_1V_1}{V_2}
\]
\[
N_2=\frac{0.100\times20.0}{25.0}
\]
\[
N_2=0.0800\,\mathrm{N}
\]
\( \textbf{Final answer:} \) The reducing-agent solution has normality \(0.0800\,\mathrm{N}\). Volumes may both be kept in \( \mathrm{mL} \) here because they appear as a ratio in the same unit.
310. A \( \mathrm{0.0500\,mol\,L^{-1}} \) \( \mathrm{K_2Cr_2O_7} \) solution is used in acidic medium. The \(n\)-factor of dichromate is \(6\). Its normality is:
ⓐ. \(0.00833\,\mathrm{N}\)
ⓑ. \(0.0500\,\mathrm{N}\)
ⓒ. \(6.00\,\mathrm{N}\)
ⓓ. \(0.300\,\mathrm{N}\)
Correct Answer: \(0.300\,\mathrm{N}\)
Explanation: \( \textbf{Molarity:} \)
\[
M=0.0500\,\mathrm{mol\,L^{-1}}
\]
\( \textbf{\(n\)-factor:} \)
\[
n=6
\]
\( \textbf{Normality relation:} \)
\[
N=M\times n
\]
\( \textbf{Calculation:} \)
\[
N=0.0500\times6
\]
\[
N=0.300\,\mathrm{N}
\]
\( \textbf{Final answer:} \) The normality is \(0.300\,\mathrm{N}\). Dividing by \(6\) would confuse normality with an equivalent-mass style calculation.
311. For \( \mathrm{H_2O_2} \) acting as an oxidising agent and forming \( \mathrm{H_2O} \), the \(n\)-factor of \( \mathrm{H_2O_2} \) is:
ⓐ. \(1\)
ⓑ. \(2\)
ⓒ. \(4\)
ⓓ. \(8\)
Correct Answer: \(2\)
Explanation: \( \textbf{Oxygen in } \mathrm{H_2O_2}: \) Oxygen is \(-1\) in peroxide.
\( \textbf{Oxygen in } \mathrm{H_2O}: \) Oxygen is \(-2\).
\( \textbf{Change per oxygen atom:} \)
\[
-1\rightarrow-2
\]
\( \textbf{Electron gain per oxygen atom:} \) \(1\) electron.
\( \textbf{Two oxygen atoms in } \mathrm{H_2O_2}: \)
\[
n=2\times1=2
\]
\( \textbf{Final answer:} \) The \(n\)-factor is \(2\). When \( \mathrm{H_2O_2} \) is reduced to water, both oxygen atoms together account for the two-electron change.
312. In another reaction, \( \mathrm{H_2O_2} \) acts as a reducing agent and forms \( \mathrm{O_2} \). What is the \(n\)-factor of \( \mathrm{H_2O_2} \) in this case?
ⓐ. \(1\)
ⓑ. \(3\)
ⓒ. \(4\)
ⓓ. \(2\)
Correct Answer: \(2\)
Explanation: \( \textbf{Starting oxygen state:} \) In \( \mathrm{H_2O_2} \), oxygen is \(-1\).
\( \textbf{Product oxygen state:} \) In \( \mathrm{O_2} \), oxygen is \(0\).
\( \textbf{Change per oxygen atom:} \)
\[
-1\rightarrow0
\]
\( \textbf{Electron loss per oxygen atom:} \) \(1\) electron.
\( \textbf{Two oxygen atoms per molecule:} \)
\[
n=2
\]
\( \textbf{Final answer:} \) The \(n\)-factor is \(2\). The role changes from oxidising agent to reducing agent, but the total electron count per \( \mathrm{H_2O_2} \) molecule is still two in this product path.
313. The statement pair below concerns \( \mathrm{H_2O_2} \).
I. \( \mathrm{H_2O_2} \) can act as an oxidising agent when it is reduced to \( \mathrm{H_2O} \).
II. \( \mathrm{H_2O_2} \) can act as a reducing agent when it is oxidised to \( \mathrm{O_2} \).
III. Its redox role is fixed and never depends on the reaction partner.
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: \( \mathrm{H_2O_2} \) contains oxygen in the intermediate oxidation state \(-1\). It can be reduced to water, where oxygen is \(-2\), so it can act as an oxidising agent. It can also be oxidised to \( \mathrm{O_2} \), where oxygen is \(0\), so it can act as a reducing agent. Statement III is false because the role depends on the reaction partner and product formed. The intermediate oxidation state allows movement in both directions.
314. A row in the following table misuses \(n\)-factor.
| Row | Reaction change | Assigned \(n\)-factor |
|---|
| P | \( \mathrm{Fe^{2+}\rightarrow Fe^{3+}} \) | \(1\) |
| Q | \( \mathrm{MnO_4^-\rightarrow Mn^{2+}} \) | \(5\) |
| R | \( \mathrm{Cr_2O_7^{2-}\rightarrow2Cr^{3+}} \) | \(6\) |
| S | \( \mathrm{H_2O_2\rightarrow H_2O} \) | \(1\) |
Which row is incorrect?
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row S
ⓓ. Row R
Correct Answer: Row S
Explanation: Row P is correct because \( \mathrm{Fe^{2+}} \) loses one electron to become \( \mathrm{Fe^{3+}} \). Row Q is correct because manganese decreases from \(+7\) to \(+2\), giving \(n=5\). Row R is correct because two chromium atoms each gain \(3\) electrons, giving \(n=6\) per \( \mathrm{Cr_2O_7^{2-}} \). Row S is incorrect because each \( \mathrm{H_2O_2} \) molecule contains two oxygen atoms changing from \(-1\) to \(-2\), so the total \(n\)-factor is \(2\). Counting only one oxygen atom gives the wrong value.
315. A \( \mathrm{0.100\,mol\,L^{-1}} \) solution of \( \mathrm{H_2O_2} \) is used in a reaction where \( \mathrm{H_2O_2\rightarrow O_2} \). The normality of the solution is:
ⓐ. \(0.0500\,\mathrm{N}\)
ⓑ. \(0.100\,\mathrm{N}\)
ⓒ. \(0.400\,\mathrm{N}\)
ⓓ. \(0.200\,\mathrm{N}\)
Correct Answer: \(0.200\,\mathrm{N}\)
Explanation: \( \textbf{Reaction role:} \) \( \mathrm{H_2O_2} \) is oxidised to \( \mathrm{O_2} \).
\( \textbf{Oxygen change:} \)
\[
-1\rightarrow0
\]
\( \textbf{Two oxygen atoms per molecule:} \)
\[
n=2
\]
\( \textbf{Molarity:} \)
\[
M=0.100\,\mathrm{mol\,L^{-1}}
\]
\( \textbf{Normality relation:} \)
\[
N=M\times n
\]
\[
N=0.100\times2
\]
\[
N=0.200\,\mathrm{N}
\]
\( \textbf{Final answer:} \) The normality is \(0.200\,\mathrm{N}\). The direction of the peroxide change must be known before applying the normality relation.
316. A graph description is given for equivalent mass of \( \mathrm{KMnO_4} \) in different media.
The x-axis lists \(n\)-factor values \(1\), \(3\), and \(5\). The y-axis shows equivalent mass for a fixed molar mass \(158\,\mathrm{g\,mol^{-1}}\). The graph decreases as \(n\) increases.
Which relation explains the graph?
ⓐ. \( \text{equivalent mass}=\frac{158}{n} \)
ⓑ. \( \text{equivalent mass}=\frac{n}{158} \)
ⓒ. \( \text{equivalent mass}=\frac{158+n}{n} \)
ⓓ. \( \text{equivalent mass}=\frac{158-n}{n} \)
Correct Answer: \( \text{equivalent mass}=\frac{158}{n} \)
Explanation: \( \textbf{Fixed molar mass:} \)
\[
M=158\,\mathrm{g\,mol^{-1}}
\]
\( \textbf{Equivalent mass relation:} \)
\[
\text{equivalent mass}=\frac{\text{molar mass}}{n}
\]
\( \textbf{Apply to } \mathrm{KMnO_4}: \)
\[
\text{equivalent mass}=\frac{158}{n}
\]
\( \textbf{Graph trend:} \) As \(n\) increases, the denominator increases, so the equivalent mass decreases.
\( \textbf{Final answer:} \) The graph is explained by \( \text{equivalent mass}=\frac{158}{n} \). This inverse relation is why acidic \( \mathrm{KMnO_4} \) has a lower equivalent mass than strongly basic \( \mathrm{KMnO_4} \).
317. In \( \mathrm{MnO_4^-} \) reductions, the product is important because:
ⓐ. it changes the atomic mass of manganese
ⓑ. it makes the charge on \( \mathrm{MnO_4^-} \) disappear before reaction
ⓒ. it fixes the manganese oxidation-number change and \(n\)-factor
ⓓ. it fixes \(n=1\) for every medium
Correct Answer: it fixes the manganese oxidation-number change and \(n\)-factor
Explanation: The starting oxidation number of manganese in \( \mathrm{MnO_4^-} \) is \(+7\). If the product is \( \mathrm{Mn^{2+}} \), the decrease is \(5\). If the product is \( \mathrm{MnO_2} \), the decrease is \(3\). If the product is \( \mathrm{MnO_4^{2-}} \), the decrease is \(1\). Thus the product controls the electron count per formula unit, while the formula \( \mathrm{MnO_4^-} \) alone is not enough.
318. An equivalent-based titration calculation says \( \mathrm{N_1V_1=N_2V_2} \). This equality is valid at equivalence because:
ⓐ. volumes of the two solutions must always be equal
ⓑ. molarities must always be equal
ⓒ. both solutions must have the same \(n\)-factor
ⓓ. oxidising and reducing equivalents are equal
Correct Answer: oxidising and reducing equivalents are equal
Explanation: At the equivalence point, the electron-donating capacity and electron-accepting capacity have matched according to the balanced redox equation. Normality already includes the \(n\)-factor, so \(NV\) gives equivalents when the same volume unit is used on both sides. The equality does not require equal volumes or equal molarities. It also does not require the two substances to have the same \(n\)-factor. Normality is useful because it converts mole amounts into electron-equivalent amounts.
319. A reducing agent \( \mathrm{R} \) has molar mass \(120\,\mathrm{g\,mol^{-1}}\). In a given reaction, one formula unit of \( \mathrm{R} \) loses \(4\) electrons. What mass of \( \mathrm{R} \) contains \(0.200\) equivalents?
ⓐ. \(3.00\,\mathrm{g}\)
ⓑ. \(12.0\,\mathrm{g}\)
ⓒ. \(6.00\,\mathrm{g}\)
ⓓ. \(24.0\,\mathrm{g}\)
Correct Answer: \(6.00\,\mathrm{g}\)
Explanation: \( \textbf{Molar mass of } \mathrm{R}: \)
\[
120\,\mathrm{g\,mol^{-1}}
\]
\( \textbf{\(n\)-factor:} \)
\[
n=4
\]
\( \textbf{Equivalent mass:} \)
\[
\frac{120}{4}=30.0\,\mathrm{g\,equiv^{-1}}
\]
\( \textbf{Given equivalents:} \)
\[
0.200\,\mathrm{equiv}
\]
\( \textbf{Mass required:} \)
\[
\text{mass}=\text{equivalents}\times\text{equivalent mass}
\]
\[
\text{mass}=0.200\times30.0
\]
\[
\text{mass}=6.00\,\mathrm{g}
\]
\( \textbf{Final answer:} \) The required mass is \(6.00\,\mathrm{g}\). Multiplying the molar mass directly by \(0.200\) would ignore that one mole supplies four equivalents.
320. A statement says, “In redox, equivalent mass can be calculated without knowing the reaction.” The best evaluation is:
ⓐ. true, because equivalent mass is always the same as molar mass
ⓑ. false; \(n\)-factor can depend on reaction and medium
ⓒ. true, because \(n\)-factor is always equal to valency
ⓓ. false, because equivalent mass is never used in redox
Correct Answer: false; \(n\)-factor can depend on reaction and medium
Explanation: Equivalent mass in redox is \( \frac{\text{molar mass}}{n} \). The molar mass is fixed for a substance, but the \(n\)-factor may change with the redox product. \( \mathrm{KMnO_4} \) is a clear example because its \(n\)-factor can be \(5\), \(3\), or \(1\) depending on the medium and product. \( \mathrm{H_2O_2} \) can also act as an oxidising or reducing agent depending on the reaction partner. Equivalent mass must therefore be tied to a stated redox change.
