Question

A galvanic cell is powered by the following redox reaction: br2 (l) 2NO(g) (l)(aq) (aq) (aq) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to decimal places.

Answer 1

Step-by-step explanation:

reduction potential of NO gas is highly unfavourable . It is - 1.7 V . So it is highly unlikely to be reduced to NO⁻ . On the other hand it is easily oxidised .

Half cell reaction of given cell

At anode ( where oxidation occurs )

NO⁻ ⇒ NO + e ( reduction potential is - 1.7 V )

At cathode ( where reduction takes place )

Br₂ + 2e ⇒2 Br⁻ ( reduction potential is 1.09 V )

(NO⁻ ⇒ NO + e ) x 2

Br₂ + 2e ⇒2 Br⁻

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Br₂ + 2NO⁻ ⇒ 2NO +2 Br⁻

Ecell = Ecathode - E anode

= 1.09 - 2 x ( - 1.7)

4.5 V