Tricky question - Redox systems in acidic and basic conditions

Watch
Serine
Badges: 1
Rep:
?
#1
Report Thread starter 5 years ago
#1
Hey guys - really stuck on this question (7d) - I'm absolutely fine with writing equations in acidic or basic conditions, I just don't understand why for this scenario why this would cause the reverse reactions to occur? Surely the same reactions should still happen? :/

Would someone mind explaining it to me?

Name:  image.jpeg
Views: 100
Size:  86.5 KBAttachment 545149545151
Attached files
0
reply
B_9710
Badges: 18
Rep:
?
#2
Report 5 years ago
#2
(Original post by Serine)
Hey guys - really stuck on this question (7d) - I'm absolutely fine with writing equations in acidic or basic conditions, I just don't understand why for this scenario why this would cause the reverse reactions to occur? Surely the same reactions should still happen? :/

Would someone mind explaining it to me?

Name:  image.jpeg
Views: 100
Size:  86.5 KBAttachment 545149545151
First you have to consider the half equation for 'redox system 7' under acidic and alkaline conditions.

In acidic conditions:
ClO-(aq) + 2H+(aq)→ ½Cl2(g) + H2O(l)

In alkaline conditions:
ClO-(aq) + H2O(l)→ ½Cl2(g) + 2OH-(aq)


Therefore under acidic conditions, the position of equilibrium lies to the right so Cl- ions reduce ClO- to Cl2, and Cl- ions are oxidised to Cl2 because Eѳ Cl2/Cl- (redox system 6) < Eѳ ClO-/Cl2 (redox system 7).

Under alkaline conditions, the position of equilibrium for 'redox system 7' lies well to the left due to the hydroxide ions. Therefore Eѳ ClO-/Cl2 (redox system 7) < Eѳ Cl2/Cl- (redox system 6) Therefore, Cl2 is reduced to ClO- ions by OH- ions. This reaction produces electrons which are then accepted by Cl2 in 'redox system 6' which reduces Cl2 to Cl- ions.

Its all to do with Le Chatelier's principle and the position of equilibrium, because that's what electrode potentials are, a measure of the position of equilibrium. If the position of equilibrium lies to the left, it has a more negative electrode potential and is a stronger reducing agent; if the position of equilibrium lies to the right, it has a more positive electrode potential and is a stronger oxidising agent.
1
reply
charco
Badges: 18
Rep:
?
#3
Report 5 years ago
#3
(Original post by B_9710)
First you have to consider the half equation for 'redox system 7' under acidic and alkaline conditions.

In acidic conditions:
ClO-(aq) + 2H+(aq)→ ½Cl2(g) + H2O(l)

In alkaline conditions:
ClO-(aq) + H2O(l)→ ½Cl2(g) + 2OH-(aq)


Therefore under acidic conditions, the position of equilibrium lies to the right so Cl- ions reduce ClO- to Cl2, and Cl- ions are oxidised to Cl2 because Eѳ Cl2/Cl- (redox system 6) < Eѳ ClO-/Cl2 (redox system 7).

Under alkaline conditions, the position of equilibrium for 'redox system 7' lies well to the left due to the hydroxide ions. Therefore Eѳ ClO-/Cl2 (redox system 7) < Eѳ Cl2/Cl- (redox system 6) Therefore, Cl2 is reduced to ClO- ions by OH- ions. This reaction produces electrons which are then accepted by Cl2 in 'redox system 6' which reduces Cl2 to Cl- ions.

Its all to do with Le Chatelier's principle and the position of equilibrium, because that's what electrode potentials are, a measure of the position of equilibrium. If the position of equilibrium lies to the left, it has a more negative electrode potential and is a stronger reducing agent; if the position of equilibrium lies to the right, it has a more positive electrode potential and is a stronger oxidising agent.
Very well explained sir.
+rep
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (18)
7.35%
Run extra compulsory lessons or workshops (37)
15.1%
Focus on making the normal lesson time with them as high quality as possible (44)
17.96%
Focus on making the normal learning resources as high quality/accessible as possible (35)
14.29%
Provide extra optional activities, lessons and/or workshops (64)
26.12%
Assess students, decide who needs extra support and focus on these students (47)
19.18%

Watched Threads

View All