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Anna007 [38]
3 years ago
13

Calculate the standard potential for a cell that employs the over-all cell reaction: 2Al(s) + 3 I2(s)→2 Al+3 + 6 I- From reducti

on potential table: Al+3 (aq) + 3e- → Al(s) -1.66 V I2(s) + 2e- → 2I- -0.54 V
Chemistry
1 answer:
Valentin [98]3 years ago
8 0

Answer:

Explanation:

2Al(s) + 3 I₂(s)   →    2 Al⁺³    +     6 I⁻

Aluminium is oxidised and iodine is reduced .

so cell potential = Ereduction - Eoxidation

Al⁺³ + 3e = Al          -  1.66 V

I₂ + 2 e = 2 I⁻             0.54 V

=  .54 - ( - 1.66 )

= 1.66 + .54

= 2.2  V

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3 years ago
Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c
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Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

Given, Moles of magnesium metal, Mg = 0.100 mol

Moles of hydrochloric acid, HCl = 0.500 mol

According to the reaction shown below:-

Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}

1 mole of Mg reacts with 2 moles of HCl

0.100 mol of Mg reacts with 2*0.100 mol of HCl

Moles of HCl must react = 0.200 mol

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Limiting reagent is the one which is present in small amount. Thus, Mg is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of Mg on reaction forms 1 mole of H_2

0.100 mole of Mg on reaction forms 0.100 mole of H_2

Mole of H_2 = 0.100 mol

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

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V is the volume

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T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K  

<u>⇒V = 2.24 L</u>

2.24 L of hydrogen gas, measured at STP, are produced.

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3 years ago
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