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3 years ago

EXPERIMENT 1: Identify the oxidation and reduction half-reactions that occur in

Chemistry

1 answer:

Hoochie [10]3 years ago

4 0

Answer:

Cathode: Mn → Mn²⁺ + 2e⁻ (Oxidation)

Anode: Zn²⁺ + 2e⁻ → Zn (Reduction)

Mn | Mn²⁺ || Zn²⁺ | Zn

Explanation:

To identify the half reaction we need to see the oxidation states.

Mn(s) → Ground state → Oxidation state = 0

Mn(NO₃)₂ → Mn²⁺ → The oxidation state has increased.

This is the oxidation reaction. It has released two electrons:

Mn → Mn²⁺ + 2e⁻

Zn(NO₃)₂ → Zn²⁺

Zn → Ground state → The oxidation state was decreased.

This is the reduction reaction. It has gained two electrons:

Zn²⁺ + 2e⁻ → Zn

Cathode: Mn → Mn²⁺ + 2e⁻

Anode: Zn²⁺ + 2e⁻ → Zn

Mn | Mn²⁺ || Zn²⁺ | Zn

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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3 years ago

A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25

mylen [45]

<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

Concentration of stock solution as 1.00 M
Volume of the stock solution as 50 mL
Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

= (1.00 M × 50 mL) ÷ 0.25 M

= 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

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kolezko [41]

Answer:

nth

Explanation:

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max2010maxim [7]

Answer:

The number of Chlorine atoms in the product is 2.

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction.

The reactants contain one chlorine molecule( $Cl_{2}$ ) which has two chlorine atoms.

Then, according to law of conservation of matter, the product must contain two chlorine atoms.

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