NO−3(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 Part A Use appropriate data to calculate E∘cell for the reaction. 3Cu(s)+2NO−3(aq)+8H+(aq)→3Cu2+(aq)+2NO(g)+4H2O(l)
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Subject:
Chemistry -
Author:
jaidyncopeland -
Created:
1 year ago
Answers 2
Answer: The standard electrode potential of the cell is 0.62 V.
Explanation:
For the given chemical equation:
3Cu(s)+2NO_3^-(aq)+8H^+(aq.)\rightarrow 3Cu^{2+}(aq.)+2NO(g)+4H_2O(l)
The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.
Oxidation half reaction: Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V ( × 3 )
Reduction half reaction: NO_3^-(aq.)+4H^+(aq.)+3e^-\rightarrow NO(g)+2H_2O(l);E^o=0.96V ( × 2 )
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the E^o_{cell} of the reaction, we use the equation:
E^o_{cell}=E^o_{cathode}-E^o_{anode}
E^o_{cell}=0.96-(0.34)=0.62V
Hence, the standard electrode potential of the cell is 0.62 V.
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