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4 years ago

Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C.

Chemistry

1 answer:

Nady [450]4 years ago

7 0

Answer:

0.895V

Explanation:

The emf of an electrochemical cell is obtained from the Nernst equation as shown in the images attached. The emf of an electrochemical cell is read from voltmeter in the circuit. The Nernst equation is used to obtain the emf of cell when it is not operated under standard conditions. That is the concentration of solutions is not 1M, temperature is not 25oc and pressure (for gases is not 1ATM)

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A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [ H 3 O ]

Mnenie [13.5K]

Answer: The hydronium ion concentration in the solution is $1.29\times 10^{-4}M$

Explanation:

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.060 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.060}{1L}\\\\\text{Molarity of HCl}=0.060M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Initial: 0.24 0.060 0.31

Final: 0.18 - 0.37

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COO^-]=0.18M$

$[C_6H_5COOH]=0.37M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.18}{0.37})\\\\pH=3.89$

To calculate the hydronium ion concentration in the solution, we use the equation:

$pH=-\log[H_3O^+]$

pH = 3.89

Putting values in above equation, we get:

$3.89=-\log[H_3O^+]$

$[H_3O^+]=10^{-3.89}=1.29\times 10^{-4}M$

Hence, the hydronium ion concentration in the solution is $1.29\times 10^{-4}M$

3 0

3 years ago

Can someone please help me:( please dont scroll create 2 quantitative observations

shusha [124]

Answer:

a quantitative observation implies that the subject can be measured by quantity, aka amount or in numbers.

Ex 1: adding one gram of salt to one gram of sugar makes two grams of seasoning. in this example, there are individual quantities (1 gram of each) and total quantity (2 grams). this only changes if the substances have a chemical reaction, such as one of them destroying the other, then the weight would change.

Ex 2: a more simple example is the weight of something. putting the substance on a scale (one specifically for whatever you are measuring, whether it be liquid or solid) is the best way to determine its quantity.

7 0

2 years ago

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate

katen-ka-za [31]

The given chemical equation is:

$Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)$

On balancing the equation we get,

$Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)$

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ $H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]$

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of $Al_{2}(SO_{4})_{3}$ reacts will be=

$15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol$

4 0

4 years ago

The ___ of an organism is its evolutionary history

harkovskaia [24]

The phylogeny of an organism is its evolutionary history.

8 0

4 years ago

Indicate whether the following balanced equations involve oxidation-reduction. Check all that apply. Check all that apply. 2H2SO

guapka [62]

Answer : The balanced equations involve oxidation-reduction are:

(a) $2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)$

(b) $3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)$

(c) $NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is:

$2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)$

This reaction involve oxidation-reduction reaction because the oxidation state bromine changes from (-1) to (0) which shows oxidation and sulfur changes from (+6) to (+4) which shows reduction.

(b) The given chemical reaction is:

$3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)$

This reaction involve oxidation-reduction reaction because the oxidation state sulfur changes from (+4) to (+6) which shows oxidation and nitrogen changes from (+5) to (+2) which shows reduction.

(c) The given chemical reaction is:

$NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)$

This reaction involve oxidation-reduction reaction because the oxidation state iodine changes from (-1) to (+5) which shows oxidation and chlorine changes from (+5) to (-1) which shows reduction.

(d) The given chemical reaction is:

$PBr_3(l)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HBr(aq)$

This reaction does not involve oxidation-reduction reaction because the oxidation state of element present on reactant and product side are same.

3 0

3 years ago