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)

The standard electrode potential of the cell is 0.62 V.

Explanation:

For the given chemical equation:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

Oxidation half reaction:       ( × 3 )

Reduction half reaction:      ( × 2 )

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the of the reaction, we use the equation:

Hence, the standard electrode potential of the cell is 0.62 V.

a is the answer to this question

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