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3 years ago

13

My coin jar has 645 quarters in it. If one Wakandan Vibranium Dollar (WVD) is worth $2.72 U.S., how many WVD can I get for my qu

arters at the bank? Give your answer to two decimal places.

1 answer:

katrin2010 [14]3 years ago

3 0

Oh yeah I don’t have internet lol lol is a little girl ugly

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2 years ago

10 points) Given the following two half-reactions, write the overall reaction in the direction in which it is product-favored, a

IceJOKER [234]

Answer:

The over all reaction :

$2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)$

The standard cell potential of the reaction is 0,.897 Volts.

Explanation:

Reduction at cathode :

$Fe^{3+}(aq)+e^-\rightarrow Fe^{2+}(s)$ ..[1]

$E^o_{Fe^{3+}/Fe^{2+}}=0.771 V$

Reduction potential of $Fe^{3+}$ to $Fe^{2+}=0.771 V$

Oxidation at anode:

$Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$ .[2]

$E^o_{Pb^{2+}/Pb}=-0.126 V$

Reduction potential of $Pb^{3+}$ to $Pb=-0.126 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=E^o_{Fe^{3+}/Fe^{2+}}-E^o_{Pb^{2+}/Pb}$

$=0.771 V-(-0.126 )=0.897 V$

The over all reaction : 2 × [1] + [2]

$2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)$

The standard cell potential of the reaction is 0,.897 Volts.

7 0

4 years ago