O ELECTROCHEMISTRY Designing a galvanic cell from two half-reactions half-reaction NO3 (ag)+4H(3eNO(g)+2 H,0 Cl2(g)+2e-? 2Cl-(aq) standard reduction potential Erd = +0.96 V red +1.359 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 Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions? ?Yes O No If you said it was possible to calculate the cell voltage, do so and enter your answer ?
Solution
Oxidation occurs at anode and reduction occurs at cathode.
At Cathode:
NO(g) + 2H2O(l) --------> NO3-(aq) + 4H+(aq) + 3e- E 0 = -0.96 V
At anode:
Cl2(g) + 2e- -------> 2Cl-(aq) E 0 = +1.359 V
Overall reaction:
NO(g) + 2H2O(l) --------> NO3-(aq) + 4H+(aq) + 3e- E 0 = -0.960 V-----------------(1)
Cl2(g) + 2e- -------> 2Cl-(aq) E 0 = +1.359 V-----------------(2)
Multiplying equation (1) by 2 and equation (1) by 3, we get;
2NO(g) + 4H2O(l) --------> 2NO3-(aq) + 8H+(aq) + 6e- E 0 = -0.960 V-----------------(3)
3Cl2(g) + 6e- -------> 6Cl-(aq) E 0 = +1.359 V-----------------(4)
Adding equation (3) and (4), we get;
2NO(g) + 4H2O(l) + 3Cl2(g) -------> 2NO3-(aq) + 8H+(aq) + 6Cl-(aq) E 0 = -0.960 V + 1.359 V
2NO(g) + 4H2O(l) + 3Cl2(g) -------> 2NO3-(aq) + 8H+(aq) + 6Cl-(aq) E 0 = 0.399 V
Yes, have enough information to calculate the cell potential and the cell potential is calculated above.
E 0 = 0.399 V
.
Salient features of Environment protection Act 1986.pptx
O ELECTROCHEMISTRY Designing a galvanic cell from two half-reactions h.docx
1. O ELECTROCHEMISTRY Designing a galvanic cell from two half-reactions half-reaction NO3
(ag)+4H(3eNO(g)+2 H,0 Cl2(g)+2e-? 2Cl-(aq) standard reduction potential Erd = +0.96 V red
+1.359 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 Write a balanced equation for the overall reaction that powers the cell. Be sure the
reaction is spontaneous as written. Do you have enough information to calculate the cell voltage
under standard conditions? ?Yes O No If you said it was possible to calculate the cell voltage, do
so and enter your answer ?
Solution
Oxidation occurs at anode and reduction occurs at cathode.
At Cathode:
NO(g) + 2H2O(l) --------> NO3-(aq) + 4H+(aq) + 3e- E 0
= -0.96 V
At anode:
Cl2(g) + 2e- -------> 2Cl-(aq) E 0
= +1.359 V
Overall reaction:
NO(g) + 2H2O(l) --------> NO3-(aq) + 4H+(aq) + 3e- E 0
= -0.960 V---------------
--(1)
Cl2(g) + 2e- -------> 2Cl-(aq) E 0
= +1.359 V--------------
---(2)
Multiplying equation (1) by 2 and equation (1) by 3, we get;
2NO(g) + 4H2O(l) --------> 2NO3-(aq) + 8H+(aq) + 6e- E 0
= -0.960 V------------
-----(3)
3Cl2(g) + 6e- -------> 6Cl-(aq) E 0
= +1.359 V-----------------(4)
2. Adding equation (3) and (4), we get;
2NO(g) + 4H2O(l) + 3Cl2(g) -------> 2NO3-(aq) + 8H+(aq) + 6Cl-(aq) E 0
= -
0.960 V + 1.359 V
2NO(g) + 4H2O(l) + 3Cl2(g) -------> 2NO3-(aq) + 8H+(aq) + 6Cl-(aq) E 0
=
0.399 V
Yes, have enough information to calculate the cell potential and the cell potential is calculated
above.
E 0
= 0.399 V
