The Mistake Bank
Chapter 8: Redox Reactions
Electron lost is electron gained... unless you miscalculate.
The Butterfly Structure Trap
Oxidation NumberScenario: Calculate the oxidation number of Chromium in $CrO_5$ (Blue Perchromate).
Student assumes normal oxide ($O^{2-}$) for all oxygens:
$$ x + 5(-2) = 0 \Rightarrow x = +10 $$
(Impossible! Max valence electrons of Cr is 6.)
Look for Peroxide Linkage!
$CrO_5$ has a butterfly structure with:
- 4 Peroxide Oxygen atoms ($-1$ each)
- 1 Normal Oxide Oxygen atom ($-2$)
$$ x + 4(-1) + 1(-2) = 0 $$
$$ x - 4 - 2 = 0 \Rightarrow \mathbf{x = +6} $$
Standard Reduction Potential
Electrochemical SeriesScenario: Given $E^\circ(Li^+/Li) = -3.05V$ and $E^\circ(Ag^+/Ag) = +0.80V$. Which is the stronger Reducing Agent?
Student thinks "Positive is better" or "Higher value is stronger".
Answer: Silver ($Ag$).
(Incorrect!)
More Negative = Stronger RA!
A highly negative Reduction Potential means the element hates being reduced and loves to oxidize (lose electrons).
Lithium ($Li$) loses electrons most easily.
Answer: $\mathbf{Li}$ is the strongest Reducing Agent.
Balancing in Basic Medium
Balancing RedoxScenario: Balance the half-reaction $MnO_4^- \rightarrow MnO_2$ in Basic medium.
Student adds $H^+$ to balance Hydrogens and leaves it there.
$$ MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O $$
($H^+$ cannot exist in Basic medium!)
Neutralize the $H^+$!
Add $OH^-$ to both sides to cancel out $H^+$.
Combine $H^+ + OH^-$ to form $H_2O$.
$$ MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- $$
KMnO4 n-Factor
TitrationScenario: Find the Equivalent weight of $KMnO_4$ in Neutral/Faintly Alkaline medium.
Student assumes n-factor is always 5 (Acidic medium value).
$$ Eq. Wt = M / 5 $$
Remember the "BAN 153" Rule!
- Basic (Strong): $Mn^{+7} \to Mn^{+6}$ (n=1)
- Acidic: $Mn^{+7} \to Mn^{+2}$ (n=5)
- Neutral: $Mn^{+7} \to Mn^{+4}$ (n=3)
Answer: $\mathbf{M/3}$
Fractional Oxidation State
StructureScenario: Calculate oxidation state of S in Tetrathionate ion ($S_4O_6^{2-}$).
Student calculates average:
$$ 4x + 6(-2) = -2 \Rightarrow x = +2.5 $$
Student thinks every sulfur atom has a +2.5 charge.
Average vs Individual!
+2.5 is just the average. The actual structure reveals:
- Two central Sulfurs (S-S bond): 0 state
- Two terminal Sulfurs ($SO_3$): +5 state
Disproportionation vs Comproportionation
Types of RedoxScenario: Identify the reaction: $NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
Student sees Nitrogen in reactant going to Nitrogen in product.
Calls it "Disproportionation" because it sounds fancy.
Check reactant states!
In $NH_4NO_3$, one N is $-3$ ($NH_4^+$) and one is $+5$ ($NO_3^-$).
They both go to $+1$ in $N_2O$.
Two states becoming one is Comproportionation (Reverse of Disproportionation).
Confess Your Sins!
"Redox is all about fair exchange. Did you give the electrons what they deserved?"
Did one of these catch you? Or do you have a different horror story from your last exam?
Scroll down to the comments section below and tell us:
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