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Mathew Mahindaratne
Mathew Mahindaratne
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Electrolysis Electrolytic Refining of Silver Nitrate

I read that the electrolytic refining of Silver is done using dilute Silver Nitratesilver nitrate as an electrolyte and a small percentage of free HNO3 $\ce{HNO3}$. The following reactions are possible at the anode:

$\ce{Ag ->Ag+ + e-}$

$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$$$\ce{Ag ->Ag+ + e-}$$ $$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does Ag$\ce{Ag}$ of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

Electrolysis of Silver Nitrate

I read that the electrolytic refining of Silver is done using dilute Silver Nitrate as an electrolyte and a small percentage of free HNO3 . The following reactions are possible at the anode:

$\ce{Ag ->Ag+ + e-}$

$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does Ag of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

Electrolytic Refining of Silver Nitrate

I read that the electrolytic refining of Silver is done using dilute silver nitrate as an electrolyte and a small percentage of free $\ce{HNO3}$. The following reactions are possible at the anode:

$$\ce{Ag ->Ag+ + e-}$$ $$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does $\ce{Ag}$ of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

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Chopin
Chopin
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I read that the electrolytic refining of Silver is done using dilute Silver Nitrate as an electrolyte and a small percentage of free HNO3 . The following reactions are possible at the anode:

$\ce{Ag+ + e- -> Ag}$$\ce{Ag ->Ag+ + e-}$

$\ce{ 4OH- -> O2 + 2H2O + 4e-}$$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does Ag of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

I read that the electrolytic refining of Silver is done using dilute Silver Nitrate as an electrolyte and a small percentage of free HNO3 . The following reactions are possible at the anode:

$\ce{Ag+ + e- -> Ag}$

$\ce{ 4OH- -> O2 + 2H2O + 4e-}$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does Ag of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

I read that the electrolytic refining of Silver is done using dilute Silver Nitrate as an electrolyte and a small percentage of free HNO3 . The following reactions are possible at the anode:

$\ce{Ag ->Ag+ + e-}$

$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does Ag of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

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Chopin
Chopin
  • 63
  • 5

Electrolysis of Silver Nitrate

I read that the electrolytic refining of Silver is done using dilute Silver Nitrate as an electrolyte and a small percentage of free HNO3 . The following reactions are possible at the anode:

$\ce{Ag+ + e- -> Ag}$

$\ce{ 4OH- -> O2 + 2H2O + 4e-}$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has $E^\circ = \pu{+0.8V}$. So why does Ag of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?