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3 years ago

A cracker crushed in water will test positive with iodine but negative with benedict’s solution. A

1 answer:

german3 years ago

7 0

A cracker which contains starch,is test positive with iodine solution but not Benedict's solution. This is because Benedict's solution is used to test for reducing sugars like glucose,galactose,fructose,maltose and lactose.In this case, the cracker is added with amylase enzyme which hydolyses starch into maltose,thus benedict's solution is test positive

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Write complete ionic equation to show the reaction of aqueous hg2(no3)2 with aqueous sodium chloride to form solid hg2cl2 and aq

Andrews [41]

Answer: Hg2(NO3)2 + 2 NaCl --------> Hg2Cl2 + 2 NaNO3 Hg2+2 + 2NO3-1 + 2Na+ Cl- ----> Hg2Cl

5 0

3 years ago

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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

What is an example of chemical change

Helga [31]

metal rusting in salt water

4 0

3 years ago

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What happens when chocolate get heated?

Leokris [45]

Answer:

Chocolate melts and becomes really sticky once heated.

Add-on:

hope that helped at all.

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2 years ago

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What processes would cause increased abrasion along the coastal region? Explain.

Andreas93 [3]

If you are talking about at the beaches waves would cause increased abrasion because the force of a wave hitting a rock or something else will slowly cause the object to wear away. Also strong winds could cause increased abrasion because the wind could pick up small sand particles that will slowly erode the thing it hits.

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3 years ago