Turn on thread page Beta
 You are Here: Home

# Chem: working out what happens in a reaction using E standard values watch

1. my method:
the redox system with the least positive/ most negative E standard value will donate electrons to the system with the most positive E standard value, because it is a better reducing agent.
its quite complicated and easy to get wrong

but apparantly there's another method involving adding the E values or something? can anyone explain it to me?

thanks
2. Yeah I think this is what you mean:

Say you have a half cell with Cu in CuSO4 and another with Zn in ZnSO4. Basically the 'Daniel Cell'.

To show in an exam what happens, you write reduction equations for both metals:
(Note that I am using a superscript 2 to represent 2+ charge, and a * symbol represents the symbol for standard conditions)

Cu² + 2e ---> Cu E* = +0.34V
Zn² + 2e ---> Zn E* = -0.76V

What you then do is re-write the most negative (aka the least oxidising) as an oxidation equation, and reverse the sign of the electrode potential.

Zn ---> Zn² + 2e E* = +0.76V
Cu² + 2e ---> Cu E* = +0.34V
------------------------------- <Adding these together...>
Zn + Cu² ---> Zn² + Cu E* = +1.1V

Little thing to note is that you sometimes get those half-equations that don't have an equal number of electrons, and to get a fully balanced ionic equation you'd need to multiply each by some number. When you get these, you don't multiply the cell potentials - they stay the same.

Some Extra Stuff:

What they may do is give you a cell in standard notation, like:
Cu | Cu² || Zn² | Zn
and say 'does this reaction work?'
You then work out the cell voltage, given by:

E*CELL = E*(Right hand side) - E*(Left hand side)
So in this case: The Zinc one - the copper one
E*CELL = -0.76 - (+0.34) = -1.1V

Since we got a big ol' negative potential it doesn't work.

If they ask you to construct a cell that does work, then always put the most negative one on the left hand side.
3. thanx that looks a lot easier than my method. but i suppose i'd be a bit hesitant to use a new method in the exam. its useful to know about anyway cos on the markschemes, theres always 2 methods, and i never know what the other ones all about.

oh and how do you do superscript?
4. i have been trying out a few questions.

so if you had

Zn2+ + 2e- <==> Zn E* -0.76
Co2+ + 2e- <==> Co E* -0.28

you would get

Zn <==> Zn2+ + 2e- E* +0.76
Co2+ + 2e- <==> Co E* -0.28
------------------------------
Zn + Co2+ --> Zn2+ + Co E* +0.48 positive therefore FEASIBLE

is this right?
5. thats right..

can anybody explain this:

say u have : Mg| Mg2+ || Fe2+| Fe (e.m.f = 1.93)

and you decrease the concentration of Mg2+..what does that do to the emf value and why?

thanks
6. Scarlet Ibis, the way I do superscript is in fact just putting Num-lock on and then, while holding Alt key type 0178 on the numpad. I only know the 4-digit code for power 2 and 3, but there are hundreds for loads of mathematical, greek and weird shape symbols. And you answered that Zn and Co question right, as Shockzz said.

Shockzz I am afraid I have no clue how changing the conc affects the E* value. All I know is that it, well, just does! I am hoping that the reason I don't know 3 days before the exam is because it's not on my syllabus.

Can I just ask what syllabuses (syllabii?) you both are doing? I'm doing AQA.
7. (Original post by Resonance)
Scarlet Ibis, the way I do superscript is in fact just putting Num-lock on and then, while holding Alt key type 0178 on the numpad. I only know the 4-digit code for power 2 and 3, but there are hundreds for loads of mathematical, greek and weird shape symbols. And you answered that Zn and Co question right, as Shockzz said.

Shockzz I am afraid I have no clue how changing the conc affects the E* value. All I know is that it, well, just does! I am hoping that the reason I don't know 3 days before the exam is because it's not on my syllabus.

Can I just ask what syllabuses (syllabii?) you both are doing? I'm doing AQA.
aqa too..its on the june 02 paper but i dont understand the markscheme.
8. i'm doing edexcel

glad i got that Q right. its such an easier method than the one i've been taught.

we dont use those cell diagram thingys so i'm not sure about the conc Q. will have a think tho.
9. shockz just learn what happens u dont need to know the explanation i dont think
10. wow ur guys way looks so hard i just use two simple methods:

OILRIG

and to calculate E cell i use anode minus cathode or reductant minus oxidant ( the word REDOX helps here as its effecticely RE-OX)

i think thats the easiest way personally
11. Shockzz I just had a go at June 02 and I don't get the mark scheme for that question either.
12. (Original post by ShOcKzZ)
thats right..

can anybody explain this:

say u have : Mg| Mg2+ || Fe2+| Fe (e.m.f = 1.93)

and you decrease the concentration of Mg2+..what does that do to the emf value and why?

thanks
Remember they are all in redox equilibria.

Mg(s)<=> Mg2+(aq) + 2e-

If Mg2+ is lowered, the equilibrium shifts to the right.

The overall reaction for this cell is Mg(s) + Fe2+(aq) ---> Mg2+(aq) + Fe(s)

Since the Mg(s)/Mg2+(aq) equilibrium shifts to the right, the overall reaction will also proceed "faster" to the right, ie the potential of proceeding from left to right is greater.

Therefore, emf (a measure of potential) increases.

Hope this helps.
selkie
13. Heeyyyy that makes sense.

Thank you Selkie.

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 26, 2004
Today on TSR

### Are you living on a tight budget at uni?

From budgets to cutbacks...

### University open days

1. University of Edinburgh
Sat, 22 Sep '18
2. University of Exeter
Undergraduate Open Days - Penryn Campus Undergraduate
Sat, 22 Sep '18
3. Loughborough University
General Open Day Undergraduate
Sat, 22 Sep '18
Poll
Useful resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE