NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l) E=0.96V ClO2(g)+e->ClO2-(aq) E=0.95V Cu2+(aq)+2e->Cu(s) E=0.34V 2H+(aq)+2e->H2(g) E=0.00V Pb2+(aq)+2e->Pb(s) E=-0.13V Fe2+(aq)+2e->Fe(s) E=-0.45V Use appropriate data to calculate E?cell for the reaction. 3Cu(s)+2NO3-(aq)+8H+(aq)->3Cu2+(aq)+2NO(g)+4H2O(l) Express your answer using two decimal places.

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Chemistry

NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l)E=0.96V
ClO2(g)+e->ClO2-(aq)E=0.95V
Cu2+(aq)+2e->Cu(s)E=0.34V
2H+(aq)+2e->H2(g)E=0.00V
Pb2+(aq)+2e->Pb(s)E=-0.13V
Fe2+(aq)+2e->Fe(s)E=-0.45V

Use appropriate data to calculate E?cell for thereaction.
3Cu(s)+2NO3-(aq)+8H+(aq)->3Cu2+(aq)+2NO(g)+4H2O(l)

Express your answer using two decimalplaces.

Answer & Explanation Solved by verified expert
3.9 Ratings (555 Votes)
Warning The E values in Tables are reduction potential 3Cus2NO3aq8Haq3Cu2aq2NOg4H2Ol The half reactions involved are Cu2aq2eCus E034V NO3aq4Haq3eNOg2H2Ol    See Answer
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