Question

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)

Answer 1

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.