Question

Based on the correct answers to the two previous questions, use the "reduction Potentials" table at the end of your textbook to determine the anode and the cathode half-cell potentials that correspond to the relevant anode and cathode half-reactions.

For the reaction of HNO2, use the following information:

NO3(-) + 3 H(+) + 2 e -> HNO2 + H2O Eo = 0.94 V.

List them in the given order with all digits shown in the tables, using the following format: "-1.23 V and 4.32 V". Note that this example is not the correct answer to the question and is given for illustration purposes only.

Answer 1

HNO2 + H2O    NO3- + 3H+ + 2e-

NO3-       HNO2    E^o_reduction  = 0.94V

  3H+     H2O     E^o_reduction  = 0V

So, H2O will be reduced and HNO2 will be oxidised

At cathode :     H2O           3H+   E^o_cathode  = 0V

At anode :    HNO2      NO3-    E^o_anode  = - 0.94V

E^o_cell =  E^o_{cathode -} E^o_anode  

   _{= 0 - ( - 0.94)}

   _{= 0.94 V}