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3 years ago

Pure water at 25°C has a pH of (1) 1 (2) 5 • (3) 7 (4) 14

Chemistry

1 answer:

Marysya12 [62]3 years ago

8 0

Answer:

7. .........................

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3 years ago

Help me answer Number 8 please

aleksandrvk [35]

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3 years ago

How many protons, neutrons, and electrons are presented in a nitrogen atom?

lina2011 [118]

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3 years ago

A chemist designs a galvanic cell that uses these two half-reactions:

Nonamiya [84]

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

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

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.

4 0

3 years ago

Question 3

aleksklad [387]

Answer:

The reverse reaction will increase.

Explanation:

If you increase amount of product of reaction, then reverse reaction will increase, and more H2 and O2 will be formed.

6 0

3 years ago