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DENIUS [597]
3 years ago
6

Оа

Chemistry
1 answer:
crimeas [40]3 years ago
7 0

Explanation:

a man out of chronium 8 = 24.6 grams

mass of oxygen = 11.8 grams

mass of oxygen = 1.2 grams

total mass =24.6 + 11.8 + 1.2

=37.6

percent composition by mass of chromium

= 24.6/37.6 = 0.654

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Answer :

(a) Reaction at anode (oxidation) : 4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-  

(b) Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O  

(c) O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}

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Explanation :

The half reaction will be:

Reaction at anode (oxidation) : Fe^{2+}\rightarrow Fe^{3+}+e^-     E^0_{anode}=+0.771V

Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O     E^0_{cathode}=+1.23V

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : 4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-     E^0_{anode}=+0.771V

Part (b):

Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O     E^0_{cathode}=+1.23V

Part (c):

The balanced cell reaction will be,

O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}

Part (d):

Now we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=(1.23V)-(0.771V)=+0.459V

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

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