[Kaleyrvt]

Observations:
i)   “A” reacts spontaneously with 1 mol/L of BNO3, 1 mol/L of “D” (NO3)2 and dilute sulfuric acid. “A” does not react with 1 mol/L of “C” (NO3)2.

ii)   “B” does not react spontaneously with any of the 1 mol/L solutions above or with dilute sulfuric acid.

iii)   “C” reacts spontaneously with dilute sulfuric acid and with 1 mol/L solution of all the other metallic salts.

iv)   “D” reacts spontaneously with 1 mol/L of BNO3 . It does not react with dilute sulfuric acid.

a)   Use the observations and arrange the following 5 reduction half-cell reactions in order, the one with the largest positive reduction potential listed first.                   

A2+   +   2e-   -->   A(s)
B+   +   e-   -->   B(s)
C2+    +   2e-   -->   C(s)
D2+   +   2e-   -->   D(s)
2H+   +   2e-   -->   H2(g)

b)   Which metal is the best reducing agent?
c)   Which ion is the best oxidizing agent?

 I get that I dont know how to figure out their potentials using the observations. I do know though that, the oxidized speciesof the half cell reaction having the highest positive reduction potential is the best oxidizing agent and the the reduced species of the half cell reaction having the most negative reduction potential is the best reducinging agent

So basically I need help getting started..I need to know how to use the information to figure out their reduction potentials. Thanks !

[Dan]

Set up a table like this:

__ A²⁺ | B⁺ | C²⁺ | D²⁺ | H⁺
A
__
B
__
C
__
D
__

Fill in the grid with a tick if the reaction happens, and a cross if it doesn't. Count the ticks for each ion. The most reactive ion will have the most ticks in its column, and will be the most easily reduced (ie. highest reduction potential).

Note: if neutral x reacts with ion yⁿ⁺, then this implies that neutral y does not react with ion x^m+

From this table you can answer the questions.
 
You could add a row for H₂ if you want, but it isn't required.

[Kaleyrvt]

highest reduction potential order: c a d b:
C2+      +   2e-  --> C(s)
A2+      +   2e-  -->A(s)
D2+      +  2e- --> D(s)
B+         +   e- -->B(s)

however, where does the 2H+   +   2e-   -->   H2(g) come in to play?

From these answers,  c is the best reducing agent and B is the best oxidizing agent ?

[DevaDevil]

however, where does the 2H+   +   2e-   -->   H2(g) come in to play?

From these answers,  c is the best reducing agent and B is the best oxidizing agent ?

Didn't you have to put the 5 half cell reactions in order? That's where the H₂ <--> 2H⁺ + 2e^- comes into play.

You are right about B being best oxidising agent and C being the best reducing agent

[Dan]

Your order is upside down. ie. from the table you should see that the B⁺ ion has the most ticks in its column (not to be confused with the B metal row), indicating that it is the most easily reduced, and hence

B⁺ + e^- ---> B    has the greatest positive reduction potential.

If you included H⁺ in your table, just use the table to slot it into place in exactly the same way as you do for the other ions.

[Kaleyrvt]

both of the replies seems to be differnt thus I am confused. Ho wdo I know what the
 2H+   + e- --> H2(g)  reacts with??? Gosh I am mixed up here
Sorry to be a pain

[Dan]

Don't worry, you're not being a pain.

Now DevaDevil said that B is the best oxidising agent (should be the B⁺ ion btw), and C is the best reducing agent. This is correct, and what I said does not conflict with it.

Let's have another look:

From the table, we can see that the B⁺ ion reacts with all the metals so for the

ion + electrons ---> metal   half reaction,

B⁺ + e^- ---> B  is the easiest, so this reduction is easiest reduction and has the greatest reduction potential.

Because B⁺ accept electrons more easily than the other ions, it is the best oxidiser (because by taking electrons from something else, it its self is reduced but the other metal is oxidised by B⁺)

The opposite is true for the C case.

C²⁺ + 2e^- ----> C    is the most difficult reduction (and has the lowest or most negative reduction potential)

This implies that the reverse reaction:

C ----> C²⁺ + 2e^-   is the easiest oxidation, and hence C is the best reducing agent.

I hope that helps

[Kaleyrvt]

Thanks dan...A big help that was!! I understand it now....so Would you put the 2H+ reaction in the middle?-I am still stuck on that one--
However your explanation on the rest was great---thanks again

[Dan]

No problem. For H⁺, use exactly the same method (put it in the table) - you will see from this that it does indeed fall right in the middle. Make sure you can complete the table including H⁺