Answered: Determine the E in each titrant…

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter10: Potentiometry And Redox Titrations

Section: Chapter Questions

Problem 8P

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Question

Determine the E in each titrant volumes: 10.00 mL; 20.00 mL; and 30.00 mL, for the titration
of 20.0 mL of 0.100 M Fe^2+ with 0.100 M Ce^4+ in a matrix of 1 M HClO4.

In 1 M HClO4:
E^o (Fe^3+/Fe^2+) = +0.767 V
E^o (Ce^4+/Ce^3+) = +1.70 V

Expert Solution

Step 1

To Find:

To find the value of $E_{c e l l}$ for the different titrant volumes.

Step 2

Explanation:

Given that the E^o value for the given half cell reactions are,

$F e^{2 +} \Leftrightarrow F e^{3 +} + e^{-} - - - E^{o} = + 0.767 V C e^{4 +} + e^{-} \Leftrightarrow C e^{3 +} - - - E^{o} = + 1.70 V$

From the E^ovalue, it shows that oxidizing agent is $C e^{4 +}$ and reducing agent is $F e^{2 +}$ .

Then, the overall cell reaction is, $F e^{2 +} + C e^{4 +} \to F e^{3 +} + C e^{3 +}$

The net emf of the cell reaction $E_{c e l l}$ can be calculated by the equation, $E_{c e l l}^{o} = E_{c a t h o d e}^{o} - E_{a n o d e}^{o}$

The potential of $E_{c a t h o d e}$ from the nernst equation is,

$E_{c a t h o d e} = E_{c e l l}^{o} - \frac{0.0591}{n} \log (\frac{[F e^{2 +}]}{[F e^{3}]}) = 0.767 - \frac{0.0591}{1} \log (\frac{[F e^{2 +}]}{[F e^{3}]})$

The $E_{c a t h o d e}$ is the potential for platinum electrosde is 0.241 V. At equivalence point, both the oxidizing and reducing reactions are in equilibrium with platinum electrode.

Step 3

Therefore, $E_{a n o d e} = E_{c a t h o d e} = E_{c e l l}^{o} - \frac{0.0591}{n} \log (\frac{[C e^{3 +}]}{[C e^{4 +}]}) = 1.70 - \frac{0.0591}{1} \log (\frac{[C e^{3 +}]}{[C e^{4 +}]})$ $2 E = 2.467 - 0.0591 \log (\frac{[C e^{3 +}] [F e^{2 +}]}{[C e^{4 +}] [F e^{3 +}]})$

The $E_{c e l l}$ after equilvalence point, $E_{c e l l}^{o} = 1.70 - 0.0591 \frac{0.0591}{n} \log (\frac{[C e^{3 +}]}{[C e^{4 +}]}) - 0.241$

Now, the potential in addition of 10.0 ml is,

Initial moles of $F e^{2 +} = 20 m l \times 0.01 M = 2 m . m o l$

Moles of $C e^{4 +} a d d e d = 10 m l \times 0.10 M = 1 m . m o l$

After the reaction,

$[F e^{3 +}] = \frac{m o l e s o f C e^{4 +}}{T o t a l v o l u m e} = \frac{1 m . m o l}{(10 + 20) m l} = 0.03 M$

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Determine the E in each titrant volumes: 10.00 mL; 20.00 mL; and 30.00 mL, for the titration of 20.0 mL of 0.100 M Fe^2+ with 0.100 M Ce^4+ in a matrix of 1 M HClO4. In 1 M HClO4: E^o (Fe^3+/Fe^2+) = +0.767 V E^o (Ce^4+/Ce^3+) = +1.70 V