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3 years ago

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Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti

on? *

1 answer:

Lina20 [59]3 years ago

6 0

<u>Answer:</u> The volume of stock solution needed is 90 mL

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=6.0M\\V_1=?mL\\M_2=1.2M\\V_2=450mL$

Putting values in above equation, we get:

$6.0\times V_1=1.2\times 450\\\\V_1=\frac{1.2\times 450}{6.0}=90mL$

Hence, the volume of stock solution needed is 90 mL

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

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

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

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

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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 $E^o_{cell}$

<u>Case 3:</u>

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

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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 $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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