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The measured cell potential, ?cell, for the cell shown is −0.3643 V at 25 ∘C. Calculate the H+ concentration.   Pt(s) | H2(g,0.869 bar) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s) The balanced reduction half-reactions for the cell and their respective standard reduction potential values, ?∘, are given.   2H+(aq)+2e−⟶H2(g)?∘Cd2+(aq)+2e−⟶Cd(s)?∘=0.00 V=−0.403

The measured cell potential, ?cell, for the cell shown is −0.3643 V at 25 ∘C. Calculate the H+ concentration.   Pt(s) | H2(g,0.869 bar) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s) The balanced reduction half-reactions for the cell and their respective standard reduction potential values, ?∘, are given.   2H+(aq)+2e−⟶H2(g)?∘Cd2+(aq)+2e−⟶Cd(s)?∘=0.00 V=−0.403

Chemistry
Chemistry
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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
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The measured cell potential, ?cell, for the cell shown is −0.3643 V at 25 ∘C. Calculate the H+ concentration.

 

Pt(s) | H2(g,0.869 bar) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s)

The balanced reduction half-reactions for the cell and their respective standard reduction potential values, ?∘, are given.

 

2H+(aq)+2e−⟶H2(g)?∘Cd2+(aq)+2e−⟶Cd(s)?∘=0.00 V=−0.403 V
To balance this equation we need to identify changes in oxidation states occurring between elements. To do this we need to remember these rules: • A neutral element on its own in its standard state has an oxidation number of zero. • The sum of the oxidation numbers for a neutral molecule must be zero. • The sum of the oxidation numbers for an ion is equal to the net charge on the ion. • In a compound or simple ion: group 1 metals are always +1, group 2 metals are always +2. • In a compound: other metals (not group 1 or 2) prefer a positive oxidation state. • In a compound: hydrogen prefers +1 (unless it is a hydride where it is -1), oxygen prefers -2, fluorine prefers -1. • In a compound with no oxygen present the other halogens will also prefer -1. The reaction is occurring in basic solution so we need to balance charge, hydrogens and oxygens with OH and H₂O. The steps are shown below: • Chromium is oxidized from zero in chromium metal to +6 in chromate anion. • Oxidation half-reaction: Cr→ CrO² + 6e- 8OH- +Cr → CrO² + 6e¯ 8OH¯ + Cr → CrO²¯ + 6e¯ + 4H₂O ● Iron is reduced from the +3 cation to the +2 cation. ● Reduction half-reaction: Fe³+ +e → Fe²+ 6Fe³+ +6e-→ 6Fe²+ • Overall reaction: 8OH- +Cr +6Fe³+ → CrO² +6Fe²+ + 4H₂O 4
Transcribed Image Text:To balance this equation we need to identify changes in oxidation states occurring between elements. To do this we need to remember these rules: • A neutral element on its own in its standard state has an oxidation number of zero. • The sum of the oxidation numbers for a neutral molecule must be zero. • The sum of the oxidation numbers for an ion is equal to the net charge on the ion. • In a compound or simple ion: group 1 metals are always +1, group 2 metals are always +2. • In a compound: other metals (not group 1 or 2) prefer a positive oxidation state. • In a compound: hydrogen prefers +1 (unless it is a hydride where it is -1), oxygen prefers -2, fluorine prefers -1. • In a compound with no oxygen present the other halogens will also prefer -1. The reaction is occurring in basic solution so we need to balance charge, hydrogens and oxygens with OH and H₂O. The steps are shown below: • Chromium is oxidized from zero in chromium metal to +6 in chromate anion. • Oxidation half-reaction: Cr→ CrO² + 6e- 8OH- +Cr → CrO² + 6e¯ 8OH¯ + Cr → CrO²¯ + 6e¯ + 4H₂O ● Iron is reduced from the +3 cation to the +2 cation. ● Reduction half-reaction: Fe³+ +e → Fe²+ 6Fe³+ +6e-→ 6Fe²+ • Overall reaction: 8OH- +Cr +6Fe³+ → CrO² +6Fe²+ + 4H₂O 4
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