A student made measurements on some electrochemical cells and calculated three q

Question

A student made measurements on some electrochemical cells and calculated three quantities The standard reaction free energy G . The equilibrium constant K at 25.0 oc The cell potential under standard conditions E His results are listed below Unfortunately, the student may have made some mistakes. Examine his results carefully and tick the box next to the incorrect quantity in each row, if any Note: If there is a mistake in a row, only one of the three quantities listed is wrong. Also, you may assume the number of significant digits in each quantity is correct Also note: for each cell, the number n of electrons transferred per redox reaction is2 calculated quantities (Check the box next to any that are wrong.) cell n 26 9.54 × 10 4.22 × 10 4.22 × 10 154. kJ/mol -0.80 V 46 -259.kl/mol 134 V O 46 259. kJ/mol O 1.34 V

Explanation / Answer

Cr in Cr+3 has oxidation state of +3

Cr in CrO4-2 has oxidation state of +6

So, Cr in Cr+3 is oxidised to CrO4-2

Fe in Fe+3 has oxidation state of +3

Fe in Fe+2 has oxidation state of +2

So, Fe in Fe+3 is reduced to Fe+2

Reduction half cell:

Fe+3 + 1e- --> Fe+2

Oxidation half cell:

Cr+3 --> CrO4-2 + 3e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

3 Fe+3 + 3e- --> 3 Fe+2

Oxidation half cell:

Cr+3 --> CrO4-2 + 3e-

Lets combine both the reactions.

3 Fe+3 + 1 Cr+3 --> 3 Fe+2 + 1 CrO4-2

Balance Oxygen by adding water

3 Fe+3 + 1 Cr+3 + 4 H2O --> 3 Fe+2 + 1 CrO4-2

Balance Hydrogen by adding H+

3 Fe+3 + 1 Cr+3 + 4 H2O --> 3 Fe+2 + 1 CrO4-2 + 8 H+

Add equal number of OH- on both sides as the number of H+

3 Fe+3 + 1 Cr+3 + 4 H2O + 8 OH- --> 3 Fe+2 + 1 CrO4-2 + 8 H+ + 8 OH-

Combine H+ and OH- to form water

3 Fe+3 + 1 Cr+3 + 4 H2O + 8 OH- --> 3 Fe+2 + 1 CrO4-2 + 8 H2O

Remove common H2O from both sides

Balanced Eqn is

3 Fe+3 + 1 Cr+3 + 8 OH- --> 3 Fe+2 + 1 CrO4-2 + 4 H2O

This is balanced chemical equation in basic medium

Answer:

3 Fe3+ + Cr3+ + 8 OH- --> 3 Fe2+ + CrO42- + 4 H2O

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