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3 years ago

At which electrode does oxidation occur in a

Chemistry

2 answers:

Verdich [7]3 years ago

7 0

Answer : The correct option is, (3) the anode in both a voltaic cell and an electrolytic cell.

Explanation :

As we know that there are two types of cell which are, electrolytic cell and voltaic cell.

Electrolytic cell : It is a cell in which the chemical reaction occurs by passing of the current from external source.

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

Voltaic 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 galvanic cell or electrochemical cell.

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

Hence, the oxidation occur in a voltaic cell and in an electrolytic cell at the anode in both a voltaic cell and an electrolytic cell.

Trava [24]3 years ago

5 0

The right answer for the question that is being asked and shown above is that: "(2) the cathode in a voltaic cell and the anode in an electrolytic cell." At the status of electrode does oxidation occur in a voltaic cell and in an electrolytic cell is that the cathode in a voltaic cell and the anode in an electrolytic cell

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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

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