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3 years ago

Which of the following substances is an element? A. carbon (C) B. water (H2O) C. methane (CH4) D. brass (Cu + Zn)

Chemistry

2 answers:

Sladkaya [172]3 years ago

8 0

The Answer is Carbon.

stealth61 [152]3 years ago

8 0

You answer is D. Brass

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Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 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

The least polar bond is found in a molecule of hi hf hcl hbr

mario62 [17]

Answer:

HI is least polar

Explanation:

5 0

3 years ago

How does heat travel from the sun to a plant in a greenhouse?

Arada [10]

Answer: Radiation

Explanation: The sun transfers heat energy through radiation since in a vacuum it does not require any medium to allow heat flow.

7 0

3 years ago

Angela is analyzing the fracture pattern of a window. She notes that the cracks stop when they intersect. Knowing this, what can

diamong [38]

Answer:

b.what type of object broke the glass

Explanation:

The point of intersection of all the fracture patterns of the window speaks to the shape, type and dimension of the object that broke the glass.

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4 years ago