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Lena [83]
3 years ago
14

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

hat species are produced at the electrodes under standard conditions? (Hint: Draw a picture of a voltaic cell)

Chemistry
1 answer:
Agata [3.3K]3 years ago
7 0

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

E^0_{[Ag^{+}/Ag]}=+0.80V

E^0_{[Cu^{2+}/Cu]}=+0.34V

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : Cu(s)\rightarrow Cu^{2+}(aq)+2e^-

Reduction half reaction (Cathode) : Ag^{+}(aq)+e^-\rightarrow Ag(s)

Thus the overall reaction will be,

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

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When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas
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<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

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When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

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Moles of methanol = 0.7605 moles

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Putting values in above equation, we get:

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