No sign flipping while figuring out the emf of voltaic cell? The Next CEO of Stack OverflowHalf cell method of voltage calculation in an electrochemical cellFinding concentrations in a voltaic cellWhy does the anode solution contain Sn2+ in a Sn-Cu voltaic cell?Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?Explain the difference in stability of permanganate ions in acidic/ alkaline solutions?At what voltage does the electrodeposition of the metal start?Positive electrode of an electrochemical cell?Do I use the Nernst equation when the concentrations of electrolyte in both half cells are equal?Does silver oxidise in a pH 1 solution?How can we directly add half cell potentials to measure the EMF of a galvanic cell?
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No sign flipping while figuring out the emf of voltaic cell?
The Next CEO of Stack OverflowHalf cell method of voltage calculation in an electrochemical cellFinding concentrations in a voltaic cellWhy does the anode solution contain Sn2+ in a Sn-Cu voltaic cell?Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?Explain the difference in stability of permanganate ions in acidic/ alkaline solutions?At what voltage does the electrodeposition of the metal start?Positive electrode of an electrochemical cell?Do I use the Nernst equation when the concentrations of electrolyte in both half cells are equal?Does silver oxidise in a pH 1 solution?How can we directly add half cell potentials to measure the EMF of a galvanic cell?
$begingroup$
I learnt that for a voltaic cell, the value for the $E_textcell^circ$ when the reaction is spontaneous is given by
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ, labeleqn:1tag1$$
so that the difference in the right gives us a positive value for $E_textcell^circ$.
But suppose we are given two half-reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation.
But when determining the $E_textcell^circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $eqrefeqn:1$ to figure out the $E_textcell^circ$? Shouldn't we do that because we reversed the second equation?
My book tells me to keep the $E_texthalf-cells^circ$ as they are written in the tables and simply put them in $eqrefeqn:1$. But why?
physical-chemistry electrochemistry redox
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
$endgroup$
add a comment |
$begingroup$
I learnt that for a voltaic cell, the value for the $E_textcell^circ$ when the reaction is spontaneous is given by
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ, labeleqn:1tag1$$
so that the difference in the right gives us a positive value for $E_textcell^circ$.
But suppose we are given two half-reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation.
But when determining the $E_textcell^circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $eqrefeqn:1$ to figure out the $E_textcell^circ$? Shouldn't we do that because we reversed the second equation?
My book tells me to keep the $E_texthalf-cells^circ$ as they are written in the tables and simply put them in $eqrefeqn:1$. But why?
physical-chemistry electrochemistry redox
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
$endgroup$
add a comment |
$begingroup$
I learnt that for a voltaic cell, the value for the $E_textcell^circ$ when the reaction is spontaneous is given by
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ, labeleqn:1tag1$$
so that the difference in the right gives us a positive value for $E_textcell^circ$.
But suppose we are given two half-reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation.
But when determining the $E_textcell^circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $eqrefeqn:1$ to figure out the $E_textcell^circ$? Shouldn't we do that because we reversed the second equation?
My book tells me to keep the $E_texthalf-cells^circ$ as they are written in the tables and simply put them in $eqrefeqn:1$. But why?
physical-chemistry electrochemistry redox
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
$endgroup$
I learnt that for a voltaic cell, the value for the $E_textcell^circ$ when the reaction is spontaneous is given by
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ, labeleqn:1tag1$$
so that the difference in the right gives us a positive value for $E_textcell^circ$.
But suppose we are given two half-reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation.
But when determining the $E_textcell^circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $eqrefeqn:1$ to figure out the $E_textcell^circ$? Shouldn't we do that because we reversed the second equation?
My book tells me to keep the $E_texthalf-cells^circ$ as they are written in the tables and simply put them in $eqrefeqn:1$. But why?
physical-chemistry electrochemistry redox
physical-chemistry electrochemistry redox
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
edited 1 hour ago
Apekshik Panigrahi
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
asked 1 hour ago
Apekshik PanigrahiApekshik Panigrahi
1365
1365
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.
answered 1 hour ago
PoutnikPoutnik
47027
$endgroup$
add a comment |
$begingroup$
Take a look at the two half reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $ceAg+(aq)$ and $ceSn^2+(aq)$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.
No sign flipping while figuring out the emf of voltaic cell?
Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ$$
If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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active
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active
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$begingroup$
The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.
answered 1 hour ago
PoutnikPoutnik
47027
$endgroup$
add a comment |
$begingroup$
The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.
answered 1 hour ago
PoutnikPoutnik
47027
$endgroup$
add a comment |
$begingroup$
The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.
answered 1 hour ago
PoutnikPoutnik
47027
$endgroup$
The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.
answered 1 hour ago
PoutnikPoutnik
47027
answered 1 hour ago
PoutnikPoutnik
47027
answered 1 hour ago
PoutnikPoutnik
47027
answered 1 hour ago
PoutnikPoutnik
47027
47027
add a comment |
add a comment |
$begingroup$
Take a look at the two half reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $ceAg+(aq)$ and $ceSn^2+(aq)$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.
No sign flipping while figuring out the emf of voltaic cell?
Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ$$
If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
$endgroup$
add a comment |
$begingroup$
Take a look at the two half reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $ceAg+(aq)$ and $ceSn^2+(aq)$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.
No sign flipping while figuring out the emf of voltaic cell?
Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ$$
If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
$endgroup$
add a comment |
$begingroup$
Take a look at the two half reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $ceAg+(aq)$ and $ceSn^2+(aq)$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.
No sign flipping while figuring out the emf of voltaic cell?
Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ$$
If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
$endgroup$
Take a look at the two half reactions:
$$
beginalign
ceAg+(aq) + e- &→ Ag(s) &qquad E^circ &= pu0.80 V \
ceSn^2+(aq) + 2 e- &→ Sn(s) &qquad E^circ &= pu-0.14 V
endalign
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $ceAg+(aq)$ and $ceSn^2+(aq)$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.
No sign flipping while figuring out the emf of voltaic cell?
Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.
$$E_textcell^circ = E_textcathode^circ - E_textanode^circ$$
If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
answered 30 mins ago
Karsten TheisKarsten Theis
3,544541
3,544541
add a comment |
add a comment |
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