The reduction half equation of the following reaction is: {Zn}_{({s})}+{CuSO}₄({aq})}{ZnSO}₄({aq})}+{Cu}_{({s})}

Correct Answer is : Cu(aq)2++2e−⟶Cu(s){Cu}_{({aq})}²+}+2e^{-}{Cu}_{({s})}Cu(aq)2++2e−⟶Cu(s)

{Zn}_{({s})}{Zn}_{({aq})}²+}+2e^{-}

{CuSO}₄({s})}+{H}₂{O}_{({l})}{Cu}_{({aq})}²+}+{SO}₄({aq})}²-}

{Cu}_{({aq})}²+}+2e^{-}{Cu}_{({s})}

{Cu}_{({aq})}²+}+e^{-}{Cu}_{({s})}

Correct Answer is : Cu(aq)2++2e−⟶Cu(s){Cu}_{({aq})}²+}+2e^{-}{Cu}_{({s})}Cu(aq)2++2e−⟶Cu(s)

Test Index

WAEC Chemistry 2023 Paper

Question : 41

Total: 50

\mathrm{Zn}_{(\mathrm{s})}+\mathrm{CuSO}_{4(\mathrm{aq})}\longrightarrow\mathrm{ZnSO}_{4(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})}

$\mathrm{Zn}_{(\mathrm{s})}\longrightarrow\mathrm{Zn}_{(\mathrm{aq})}^{2+}+2e^{-}$
$\mathrm{CuSO}_{4(\mathrm{s})}+\mathrm{H}_2\mathrm{O}_{(\mathrm{l})}\longrightarrow\mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}$
$\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2e^{-}\longrightarrow\mathrm{Cu}_{(\mathrm{s})}$
$\mathrm{Cu}_{(\mathrm{aq})}^{2+}+e^{-}\longrightarrow\mathrm{Cu}_{(\mathrm{s})}$

Solution:

The reduction half equation for the given reaction is:

\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2e^{-}\longrightarrow\mathrm{Cu}_{(\mathrm{s})}

In the overall reaction, copper(II) ions (

\mathrm{Cu}^{2+}

) gain two electrons (

2e^{-}

) and get reduced to form solid copper ( Cu ) metal. This is the reduction half-reaction in the redox reaction between zinc

(\mathrm{Zn})

and copper(II) sulfate

(\mathrm{CuSO}_4)

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