Study Chemistry notes on Redox Equations for Secondary. Useful for revision, homework, and exam preparation in Zimbabwe.
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Chemguide - answers
REDOX EQUATIONS
1. a) The reaction between chlorine gas and bromide ions:
\[
\begin{aligned}
\mathrm{Cl}_{2}+2 \mathrm{e}^{-} | \longrightarrow 2 \mathrm{Cl}^{-}
2 \mathrm{Br}^{-} | \longrightarrow \mathrm{Br}_{2}+2 \mathrm{e}^{-}
\end{aligned}
\]
This is easy because two electrons are involved in both half-equations. All you need to do is add the two equations together to give
\[
\mathrm{Cl}_{2}+2 \mathrm{Br}^{-} \longrightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_{2}
\]
b) The reaction between iron(II) ions and acidified potassium manganate(VII) solution:
\[
\begin{aligned}
\mathrm{Fe}^{2+} | \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}
\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} | \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}
\end{aligned}
\]
This time, you would need to multiply the iron half-reaction by 5 in order to produce the 5 electrons needed by the second half-reaction. Then add them together to give
\[
5 \mathrm{Fe}^{2+}+\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}
\]
c) The reaction between ethanol and acidified potassium dichromate(VI) solution to give ethanal:
\[
\begin{aligned}
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} | \longrightarrow \mathrm{CH}_{3} \mathrm{CHO}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} | \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\end{aligned}
\]
To get the 6 electrons needed for the second half-reaction, you would need to multiply the first one by 3 . Then add them together:
\[
\begin{aligned}
3 \times\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right. | \left.\longrightarrow \mathrm{CH}_{3} \mathrm{CHO}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\right)
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} | \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\end{aligned}
\]
That leaves you with:
\[
3 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+} \longrightarrow 3 \mathrm{CH}_{3} \mathrm{CHO}+6 \mathrm{H}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\]
... but there are hydrogen ions on both sides of the equation. Simplify it to give
\[
3 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+8 \mathrm{H}^{+} \longrightarrow 3 \mathrm{CH}_{3} \mathrm{CHO}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} …