SEMI FINAL STAGE 2025
Mfantsipim School: 42 points
GSTS: 23 points
Mankranso SHS: 15 points
QUESTION
Determine the number of electrons needed to balance the half-reaction for the reduction of $MnO_4^-$ in basic medium.
ANSWER: 3 electrons.
SOLUTION:
In basic/neutral medium, permanganate is reduced to manganese dioxide: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
The shortcut is tracking only the oxidation state change of Mn: it goes from $+7$ (in $MnO_4^-$) to $+4$ (in $MnO_2$), a change of 3, so exactly 3 electrons are gained.
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PRACTICE QUESTIONS
1. QUESTION: Determine the number of electrons needed to balance the half-reaction for the reduction of $MnO_4^-$ to $MnO_4^{2-}$.
ANSWER: 1 electron
SOLUTION:
Mn goes from +7 to +6, a change of 1: $MnO_4^-+e^-\rightarrow MnO_4^{2-}$.
2. QUESTION: Determine the number of electrons needed to balance the half-reaction for the reduction of one $O_2$ molecule to $OH^-$ in basic medium.
ANSWER: 4 electrons
SOLUTION:
$O_2+2H_2O+4e^-\rightarrow4OH^-$: each O goes from 0 to −2.
3. QUESTION: Determine the number of electrons needed to balance the half-reaction for the reduction of $ClO^-$ to $Cl^-$ in basic medium.
ANSWER: 2 electrons
SOLUTION:
Cl goes from +1 to −1: $ClO^-+H_2O+2e^-\rightarrow Cl^-+2OH^-$.
4. QUESTION: Determine the number of electrons needed to balance the half-reaction for the reduction of $Fe^{3+}$ to $Fe^{2+}$.
ANSWER: 1 electron
SOLUTION:
Fe goes from +3 to +2: $Fe^{3+}+e^-\rightarrow Fe^{2+}$.
5. QUESTION: Determine the number of electrons needed to balance the half-reaction for the reduction of $NO_3^-$ to $NO_2$ in acidic medium.
ANSWER: 1 electron
SOLUTION:
N goes from +5 to +4: $NO_3^-+2H^++e^-\rightarrow NO_2+H_2O$.
6. QUESTION: Determine the number of electrons needed to balance the half-reaction for the reduction of one $Cl_2$ molecule to $Cl^-$.
ANSWER: 2 electrons
SOLUTION:
Each Cl goes from 0 to −1: $Cl_2+2e^-\rightarrow2Cl^-$.