Question

half-reaction standard reduction potential C1 (9)+2e" - 21 (aq) NO, (aq)+41 (aq)+3e" - NO(g)+2 H,00 od = +1.359 V Ered = +0.96 V Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. Write a balanced equation for the half-reaction that happens at the anode. I Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written Yes Do you have enough information to calculate the cell voltage under standard conditions? No If you said it was possible to calculate the cell voltage, do so and enter your answer here. Be sure your answer has the er v

Answer 1

Half reaction at cathode( Reduction):

$Cl_2(g)+2e^- \rightarrow2Cl^-(aq)$

------->(a)

Half reaction at cathode( Oxidation):

$NO(g)+2H_2O(l) \rightarrow NO_3^-(aq)+4H^+(aq)+3e^-$

--------->(b)

Overall balanced Reaction:

3(a)+2(b)=>

$3Cl_2(g)+2NO(g)+4H_2O(l) \rightarrow 6Cl^-(aq)+2NO_3^-(aq)+8H^+(aq)$

Yes it is possible to calculate standard reduction potential.

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$=>E^o_{cell}=1.359-0.96V$

  $=0.399V$