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julsineya [31]
2 years ago
12

I2 + KOH = KIO3 + KI + H2O Marque la(s) respuesta(s) falsas: La suma de coeficientes mínimos del agua y el agente reductor es 6

El KI es la forma reducida El KOH es el agente reductor La suma de electrones transferidos más el coeficiente mínimo del agua suman 16 La relación entre el agente oxidante y el agente reductor es 1
Engineering
1 answer:
True [87]2 years ago
8 0

Answer:

Las declaraciones falsas incluyen

- El KOH es el agente reductor.

- La suma de electrones transferidos más el coeficiente mínimo de agua suman 16.

Todas las otras declaraciones son ciertas.

The false statements include

- The KOH is the reducing agent.

- The sum of transferred electrons plus the minimum coefficient of water add up to 16.

All the other statements are true.

Explanation:

Es evidente que esta es una reacción redox en presencia de medio básico. Entonces, equilibraremos esta reacción redox en pasos. I₂ + KOH → KIO₃ + KI + H₂O

Paso 1 Eliminar los iones espectadores; Estos son los iones que aparecen en ambos lados de la reacción. Es evidente que solo el ion de potasio (K⁺) es el ion espectador de esta reacción.

I₂ + OH⁻ → IO₃⁻ + I⁻ + H₂O

Paso 2

Separamos la reacción en las medias reacciones de oxidación y reductina. La oxidación es la pérdida de electrones que conduce a un aumento del número de oxidación del ion, mientras que la reducción es la ganancia de elecrones que conduce a una disminución en el número de oxidación del ion. También es evidente que es el gas de yodo el que se reduce y oxida para esta reacción.

El gas de yodo se reduce a I⁻ (el número de oxidación se reduce de 0 a -1) y el gas de yodo se oxida a IO₃⁻ (el número de oxidación de yodo aumenta de 0 en gas de yodo a +5 en IO₃⁻)

Reducción media reacción

I₂ → I⁻

Media reacción de oxidación

I₂ + OH⁻ → IO₃⁻ + H₂O

Paso 3

Equilibramos las medias reacciones y agregamos los respectivos electrones transferidos

Reducción media reacción

I₂ → 2I⁻

I₂ + 2e⁻ → 2I⁻

Media reacción de oxidación

I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O

I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻

Paso 4

Balancee el número de electrones en las dos medias reacciones

[I₂ + 2e⁻ → 2I⁻] × 5

[I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻] × 1

5I₂ + 10e⁻ → 10I⁻

I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻

Paso 5

Agregue las dos medias reacciones y elimine cualquier especie que aparezca en ambos lados

5I₂ + 10e⁻ + I₂ + 12OH⁻ → 10I⁻ + 2IO₃⁻ + 6H₂O + 10e⁻

Entonces, eliminamos los 10 electrones que fueron transferidos en la reacción balanceada

6I₂ + 12OH⁻ → 10I⁻ + 2IO₃⁻ + 6H₂O

Paso 6

Reintroducimos la especie eliminada desde el principio (el ion potasio)

6I₂ + 12KOH → 10KI + 2KIO₃ + 6H₂O

Los coeficientes mínimos son entonces

3I₂ + 6KOH → 5KI + KIO₃ + 3H₂O

Luego verificamos cada una de las declaraciones proporcionadas para elegir las falsas.

- La suma de los coeficientes mínimos del agua y el agente reductor es 6.

El gas yodo es el agente reductor y oxidante. Coeficiente mínimo de agua y gas de yodo = 3 + 3 = 6 Esta afirmación es cierta.

- El KI es la forma reductora KI resulta de la semirreacción de reducción.

Por lo tanto, es la forma reducida del gas de yodo. Esta afirmación es cierta. - El KOH es el agente reductor. KOH no es el agente reductor. Esta afirmación es falsa.

- La suma de los electrones transferidos más el coeficiente mínimo de agua suman 16.

Electrones transferidos = 10

Coeficiente mínimo de agua = 3

Suma = 13 y no 16.

Esta afirmación es falsa.

- La proporción del agente oxidante y el agente reductor es 1.

Dado que el gas yodo es el agente reductor y oxidante, la proporción de estos dos es verdaderamente 1. Esta afirmación es cierta.

¡¡¡Espero que esto ayude!!!

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