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2 years ago

Balance the following Equation:

Chemistry

1 answer:

nekit [7.7K]2 years ago

8 0

Answer:

Zn is the excess reactant (0.079 moles in excess)

Option C is correct

Explanation:

Step 1: Data given

Mass of Zinc (Zn) = 50.0 grams

Mass of Hydrogen chloride (HCl) = 50.0 grams

atomic mass Zn = 65.38 g/mol

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol

Moles Zn = 0.764 moles

Moles HCl = 50.0 grams / 36.46 g/mol

Moles HCl = 1.37 moles

Step 4: Calculate limiting reactant

For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Zn is in excess. There will react 1.37/2 = 0.685 moles

There will be 0.764 -0.685 = 0.079 moles in excess

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antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

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