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2 years ago

Which Ions oxidize aluminum? Li^+, Ca^2+, Ag+, Sn^2+

Chemistry

1 answer:

pishuonlain [190]2 years ago

8 0

Aluminum can be oxidized by $Li^{+}$ and $Ca^{2+}$

<h3>What is oxidizing?</h3>

Some of the metal corrodes (or oxidizes) and forms the corresponding metal oxide on the surface as a result of a chemical reaction between the metal surface and the oxygen in the air. The corrosion products that occur in some metals, like steel, are highly apparent and loose.

According to reactivity series (The array of metals in the descending order of their reactivities is referred to as the metals' reactivity series. It is sometimes referred to as the metals in the activity series.)

Lithium and calcium ions are more reactive than aluminum ion and they are less electronegative.

Since silver and tin are more electronegative than aluminum so, they cannot oxidize aluminum.

∴ $Li^{+}$ and $Ca^{+2}$ ions can oxidize aluminum.

Learn more about electronegativity here brainly.com/question/16446391

#SPJ10

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Answer:

water and oxygen

Explanation:

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3 years ago

Explain how lines latitude and longitude help people find locations on earth.

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3 years ago

A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$

(b) Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$

(c) $O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : $Fe^{2+}\rightarrow Fe^{3+}+e^-$ $E^0_{anode}=+0.771V$

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : $4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-$ $E^0_{anode}=+0.771V$

Part (b):

Reaction at cathode (reduction) : $O_2+4H^++4e^-\rightarrow 2H_2O$ $E^0_{cathode}=+1.23V$

Part (c):

The balanced cell reaction will be,

$O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}$

Part (d):

Now we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=(1.23V)-(0.771V)=+0.459V$

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

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