How do you write an equation for “Aluminium + copper $(II)$ sulphate → copper +aluminium sulphate”?
Answer
624k+ views
Hint:Chemical equation is defined as the representation of chemical reaction. The reactant and the product are represented with their respective chemical formula. In this reaction, symbols are also used to represent the physical state of the substance and the direction of the reaction.
Complete answer:
Reactivity series is defined as the series of metals in which the metals are arranged in the order of reactivity from highest to lowest. Let us take a general representation of the displacement reaction.
${{A}_{(s)}}+{{B}^{+}}_{(aq)}\to {{A}^{+}}_{(aq)}+{{B}_{(s)}}$
Metal $A$ replaces metal $B$ only when the reactivity of metal \[A\] is higher than the reactivity of \[B\] in the series.
In this question, a displacement reaction is given where the aluminium lies above than the copper in the reactivity series, therefore, aluminium displaces copper.
In this reaction, the blue color of copper sulphate decolorizes and turns into colorless.
In the given reaction, the electrons are transferred from aluminium and get oxidized and copper gets reduced.
Oxidation of aluminium:
$A{{l}_{(s)}}\to A{{l}^{3+}}_{(aq)}+3{{e}^{-}}$
Reduction of copper:
$C{{u}^{2+}}_{(aq)}+2{{e}^{-}}\to C{{u}_{(s)}}$
The net reaction can be written as follows:
$\left[ Al\to A{{l}^{3+}}+3{{e}^{-}} \right]\times 2$
$\left[ C{{u}^{2+}}+2{{e}^{-}}\to C{{u}_{(s)}} \right]\times 3$
The reaction can be combined as follows:
$2A{{l}_{(s)}}+3C{{u}^{2+}}_{(aq)}\to 2A{{l}^{3+}}_{(aq)}+3C{{u}_{(s)}}$
Hence, the equation will be
$2A{{l}_{(s)}}+3CuS{{O}_{4}}_{(aq)}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}_{(aq)}+3C{{u}_{(s)}}$
Note:For representing the direction of reaction different symbols are used:
-For forward reaction, the symbol used is $'\to '$
-For an equilibrium reaction, the symbol used is $'\rightleftharpoons '$
-For both forward reaction and backward reaction, the symbol used is $'\rightleftarrows '$
For representing the physical state of the substances:
-Symbol used for solid state is $(s)$
-Symbol used for liquid state is $(l)$
-Symbol used for gaseous state is $(g)$
-Symbol used for aqueous state is $(aq)$
Complete answer:
Reactivity series is defined as the series of metals in which the metals are arranged in the order of reactivity from highest to lowest. Let us take a general representation of the displacement reaction.
${{A}_{(s)}}+{{B}^{+}}_{(aq)}\to {{A}^{+}}_{(aq)}+{{B}_{(s)}}$
Metal $A$ replaces metal $B$ only when the reactivity of metal \[A\] is higher than the reactivity of \[B\] in the series.
In this question, a displacement reaction is given where the aluminium lies above than the copper in the reactivity series, therefore, aluminium displaces copper.
In this reaction, the blue color of copper sulphate decolorizes and turns into colorless.
In the given reaction, the electrons are transferred from aluminium and get oxidized and copper gets reduced.
Oxidation of aluminium:
$A{{l}_{(s)}}\to A{{l}^{3+}}_{(aq)}+3{{e}^{-}}$
Reduction of copper:
$C{{u}^{2+}}_{(aq)}+2{{e}^{-}}\to C{{u}_{(s)}}$
The net reaction can be written as follows:
$\left[ Al\to A{{l}^{3+}}+3{{e}^{-}} \right]\times 2$
$\left[ C{{u}^{2+}}+2{{e}^{-}}\to C{{u}_{(s)}} \right]\times 3$
The reaction can be combined as follows:
$2A{{l}_{(s)}}+3C{{u}^{2+}}_{(aq)}\to 2A{{l}^{3+}}_{(aq)}+3C{{u}_{(s)}}$
Hence, the equation will be
$2A{{l}_{(s)}}+3CuS{{O}_{4}}_{(aq)}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}_{(aq)}+3C{{u}_{(s)}}$
Note:For representing the direction of reaction different symbols are used:
-For forward reaction, the symbol used is $'\to '$
-For an equilibrium reaction, the symbol used is $'\rightleftharpoons '$
-For both forward reaction and backward reaction, the symbol used is $'\rightleftarrows '$
For representing the physical state of the substances:
-Symbol used for solid state is $(s)$
-Symbol used for liquid state is $(l)$
-Symbol used for gaseous state is $(g)$
-Symbol used for aqueous state is $(aq)$
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