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The given redox reaction is: Cr2O72-aq + C2O42-aq → Cr3+aq + CO2g Oxidation state of…

Answered: acıdıc 2- 3+ + CO2 (5) 2- basic

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Expert Solution

Step 1: Reaction - 1 in acidic medium

The given redox reaction is:

$C r_{2} {O_{7}}^{2 -}_{(a q)} + C_{2} {O_{4}}^{2 -}_{(a q)} \to C {r^{3 +}}_{(a q)} + C O_{2_{(g)}}$

Oxidation state of Cr in Cr₂O₇^2-:

$(2 \times O x i d a t i o n s t a t e o f C r) + (- 2 (c h a r g e o f O a t o m) x 7) = - 2 O x i d a t i o n s t a t e o f C r = + 6$

Oxidation state of C in C₂O₄^2- :

$(2 \times O x i d a t i o n s t a t e o f C) + (- 2 (c h a r g e o f O a t o m) x 4) = - 2 O x i d a t i o n s t a t e o f C = + 3$

Oxidation state of C in CO₂ :

$(1 \times O x i d a t i o n s t a t e o f C) + (- 2 (c h a r g e o f O a t o m) x 2) = 0 O x i d a t i o n s t a t e o f C = + 4$

Therefore Chromium is undergoing reduction and Carbon is undergoing Oxidation.

Step 2

The reduction half reaction is:

$C r_{2} {O_{7}}^{2 -}_{(a q)} \to C {r^{3 +}}_{(a q)}$

Balancing Cr atoms:

There are 2 Cr atoms on reactant side and 1 Cr atom in products side. Multiplying Cr³⁺ with 2 will give equal number of Cr atoms on both the sides. The reaction changes to:

$C r_{2} {O_{7}}^{2 -}_{(a q)} \to 2 C {r^{3 +}}_{(a q)}$

Balancing O atoms:

There are 7 O atoms on reactant side and zero O atom in products side. Adding 7 H₂O on products side. The reaction changes to:

$C r_{2} {O_{7}}^{2 -}_{(a q)} \to 2 C {r^{3 +}}_{(a q)} + 7 H_{2} O$

Balancing H atoms:

There are zero H atoms on reactant side and 14 H atoms in products side. Addition of 14 H⁺ to reactants side will give equal number of H atoms on both the sides. The reaction changes to:

$C r_{2} {O_{7}}^{2 -}_{(a q)} + 14 H^{+} \to 2 C {r^{3 +}}_{(a q)} + 7 H_{2} O$

Balancing the charge:

There is +12 charge on reactant side and +3 charge on products side. Addition of 9 electrons to products side will balance the charge. The reaction changes to:

$C r_{2} {O_{7}}^{2 -}_{(a q)} + 14 H^{+} + 9 e^{-} \to 2 C {r^{3 +}}_{(a q)} + 7 H_{2} O$

Step 3

The Oxidation half reaction is:

$C_{2} {O_{4}}^{2 -}_{(a q)} \to C O_{2_{(g)}}$

Balancing C atoms:

There are 2 C atoms on reactant side and 1 C atom on products side. Multiplying CO₂ with 2 will give equal number of C atoms on both the sides. The reaction changes to:

$C_{2} {O_{4}}^{2 -}_{(a q)} \to 2 C O_{2_{(g)}}$

Balancing the charge:

There is -2 charge on reactant side and zero charge on products side. Addition of 2 electrons to products side will balance the charge. The reaction changes to:

$C_{2} {O_{4}}^{2 -}_{(a q)} \to 2 C O_{2_{(g)}} + 2 e^{-}$

Now balancing the electrons in both the half reactions:

Multiplying reduction half reaction with 2 gives:

$2 \times (C r_{2} {O_{7}}^{2 -}_{(a q)} + 14 H^{+} + 9 e^{-} \to 2 C {r^{3 +}}_{(a q)} + 7 H_{2} O) 2 C r_{2} {O_{7}}^{2 -}_{(a q)} + 28 H^{+} + 18 e^{-} \to 4 C {r^{3 +}}_{(a q)} + 14 H_{2} O$

Multiplying oxidation half reaction with 9 gives:

$9 \times (C_{2} {O_{4}}^{2 -}_{(a q)} \to 2 C O_{2_{(g)}} + 2 e^{-}) 9 C_{2} {O_{4}}^{2 -}_{(a q)} \to 18 C O_{2_{(g)}} + 18 e^{-}$

Now adding both the half reactions gives:

$2 C r_{2} {O_{7}}^{2 -}_{(aq)} + 28 H^{+} + 18 e^{-} \to 4 C {r^{3 +}}_{(aq)} + 14 H_{2} O (+) 9 C_{2} {O_{4}}^{2 -}_{(aq)} \to 18 C O_{2_{(g)}} + 18 e^{-} - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 C r_{2} {O_{7}}^{2 -}_{(aq)} + 28 H^{+} + 18 e^{-} + 9 C_{2} {O_{4}}^{2 -}_{(aq)} \to 4 C {r^{3 +}}_{(aq)} + 14 H_{2} O + 18 C O_{2_{(g)}} + 18 e^{-}$

Therefore the balanced redox reaction in acidic medium is:

$2 C r_{2} {O_{7}}^{2 -}_{(aq)} + 28 H^{+} + 9 C_{2} {O_{4}}^{2 -}_{(aq)} \to 4 C {r^{3 +}}_{(aq)} + 14 H_{2} O + 18 C O_{2_{(g)}}$

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