You are titrating 100.0 mL of 0.0200 M Fe3 in 1 M HClO4 with 0.100 M Cu to give

Question

You are titrating 100.0 mL of 0.0200 M Fe3 in 1 M HClO4 with 0.100 M Cu to give Fe2 and Cu2 using Pt and saturated Ag | AgCl electrodes to find the endpoint. Please answer all parts of the question (a, b, and c)

(a) Write the balanced titration reaction.

(b) Complete the two half reactions for the Pt electrode.

(c) Calculate the value of E for the cell when at the volume 21.0 mL of the Cu titrant has been added: (Activity coefficients may be ignored as they tend to cancel when calculating concentration ratios.)

Explanation / Answer

To find the balance reaction of part a, we can start with part b.So let me start with part b.

part b: Fe3+ + e- --> Fe2+

Multiply the above eqaution by 2,we will get

2Fe3+ + 2e- --> Fe2+ .....(i)

and Cu --> Cu2+ + 2e- ......(ii)

These equations are the answer of part b

Now if we add both the equations, we will get

answer of part a, that is

Part a 2Fe3+ + Cu --> 2Fe2+ + Cu2+

Part c

We will have to use Nernst Equation for Copper

Nernst equation is given by

E= E0 - (0.059/n) log ([product]/[reactant])

n is number of electrons transferring, in our case it is 2.

E0 for copper is not given, so let it be E0

Now concentration of product on adding 21ml of titrant will be 21-(21*0.100) = 21-2.1=18.9

and concentration of reactant will be 21*0.100= 2.1

Putting values in Nernst equation

E=E0 - (0.059/2) log (18.9/2.1) = E0 - 0.0287

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