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Half-cell reactions - Reduction potential

Submitted by Kayla on Sun, 11/29/2009 - 19:21

i) A reacts spontaneously with 1 mol/L BNO3, 1 mol/L D(NO3)2 and dilute sulfuric acid. A does not react with 1 mol/L 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 solutions of all the other metallic salts.
iv) D reacts spontaneously with 1 mol/L BNO3. It does not react with dilute sulfuric acid.

a) Use the observations and arrange the following five 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 don't understand these observations at all..

These are displacement reactions (which you might call replacement reactions)

eg Mg will displace Cu from Cu(NO3)2
Mg + Cu(NO3)2 ----> Mg(NO3)2 + Cu

This will happen when a metal higher in the reactivity series (Mg) displaces a metal lower in the series (Cu) from its salt. Also in the reactivity series only metals higher than H can displace H from an acid. Have a look at
for details on the chemical order of activity

So from 1) I deduce that A is more reactive than B and D but not C. I also know that A is higher in the series than hydrogen as it displaces hydrogen from an acid    A + H2SO4 ----> ASO4 + H2. So the order is C,A ( with B,D in no particular order)
ii) This tell me B is the least reactive of all and is below H in the activity series. So C,A,D,B is the order
iii) this confirms that C is the most reactive and high up in the  series eg Mg perhaps
iv) D is more reactive that B but is lower than H so could be Cu while B might be Ag or Au

2H+ + 2e- ? H2(g)  is a reference half cell in the electrochemical series and has a value of 0.00v

Less reactive metals will have a more positive value while more reactive metals will have a negative value. Have a look at

Based on the order C, A, (H), D, B.
So B will be the most positive, followed by D, H, A and C

B+ + e- ? B(s)  most positive Eo (perhaps Ag)
D2+ + 2e- ? D(s)  positive E0  ( perhaps Cu)
2H+ + 2e- ? H2(g) Eo = 0.00v
A2+ + 2e- ? A(s)  negative E0  (perhaps Zn)
C2+ + 2e- ? C(s)  most negative Eo  (perhaps Mg)

In the electrochemical series these would normally be written in the reverse order with the  most negative values at the top and the most positive values at the bottom

As for the metal which the best reducing agent, you want the one with the least positive value eg  C will lose electrons most readily and behave as a reducing agent.

The best oxidising agent is the ion with the most positive Eo

Thank you so much for your help, very much appreciated! Makes much more sense now.