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2 years ago

Help me please Answer & exp.

Chemistry

1 answer:

nordsb [41]2 years ago

8 0

Explanation:

Some Rules Regarding Oxidation Numbers:

- Hydrogen has oxidation number of + 1 except in hydrides where it is -1

- Oxygen has oxidation number of -2 except in peroxides where it is -1

- Some elements have fixed oxidation numbers. E.g Halogen group elements has oxidation number of -1

- Oxidation number of a compound is the sum total of the individual elements and a neutral compound has oxidation number of 0.

A. HI

Hydrogen has oxidation of + 1

Oxidation number of I:

1 + x = 0

x = -1

B. PBr3

Br has oxidation number of - 1

Oxidation number of Pb:

x + 3 (-1) = 0

x = + 3

C. KH

Hydrogen has oxidation of + 1

Oxidation number of K:

1 + x = 0

x = -1

D. H3PO4

Hydrogen has oxidation number of + 1

Oxygen has oxidation number of -2

Oxidation number of P:

3(1) + x + 4(-2) = 0

3 + x - 8 =0

x = 5

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Explanation:

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2 years ago

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Olegator [25]

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3 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

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

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

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