How do you write a equation for "Aluminum + copper (ll) sulfate -> copper + aluminum sulfate"?

Answer 1

#2Al(s)+3CuSO_4(aq)->Al_2(SO_4)_3(aq)+3Cu(s)#

The following response would occur:

#2Al(s)+3CuSO_4(aq)->Al_2(SO_4)_3(aq)+3Cu(s)#

This is an oxidation-reduction reaction were electrons are transferred from Aluminum #Al# which get oxidized to Copper #Cu^(2+)# which gets reduced.

Oxidation: #Al(s)->Al^(3+)(aq)+3e^-#

Reduction: #Cu^(2+)(aq)+2e^(-)->Cu(s)#

In that case, the net redox reaction would be:

#color(red)(2xx)(Al(s)->Al^(3+)(aq)+cancel(3e^-))# #color(red)(3xx)(Cu^(2+)(aq)+cancel(2e^(-))->Cu(s))# #---------------# #2Al(s)+3Cu^(2+)(aq)->2Al^(3+)(aq)+3Cu(s)#

Note that #SO_4^(2-)# is a spectator ion.

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Answer 2

Kai Bowman

The balanced chemical equation for the reaction between aluminum and copper (II) sulfate is:

2Al + 3CuSO₄ → 3Cu + Al₂(SO₄)₃

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you write a equation for "Aluminum + copper (ll) sulfate -&gt; copper + aluminum sulfate"?

How do you write a equation for "Aluminum + copper (ll) sulfate -> copper + aluminum sulfate"?