# Balance Redox Reaction/Equation

7.1 What is the overall reaction (equation):
Cr2O7-2  +  14H+  + 6e-   -->   2Cr+3  +  7H2O
H2S   -->   S  +  2H+  +  2e-

Solution:
Insert following equations in the half cells:
Cr2O7-2  +  14H+ +  6e-  >  2Cr+3  +  7H2O
H2S  >  S  +  2H+  + 2e-
and calculate.

7.2 What is the overall reaction (equation):
Cr  -->  Cr+3  +  3e-
MnO4- +  8H+  + e-  -->  Mn+2 +  4H2O

Solution:
Insert following half cells reactions:
Cr  >  Cr+3  +  3e-
MnO4- +  8H+  +  e-  >  Mn+2 + 4H2O
and calculate.

7.3 Balance: OH- + ClO- + S2O3-2 --> Cl- + SO4-2 + H2O

Solution:
Insert following equation i one of the half cell fields:
OH- + ClO- + S2O3-2 > Cl- + SO4-2 + H2O
and calculate.

7.4 Dropping a piece of sodium in water. What is the overall reaction. Is this a spontaneous reaction, explain?
2H2O + 2e- --> H + 2OH-       E0 = -0.83 V
Na + e --> Na                         E0 = -2.71 V
Solution:
Insert following half cells reactions and potentials:
2H2O + 2e-   >  H + 2OH-
Na+  +  e-     > Na
Remember to turn second half cell reaction.
Calculate.
Spontaneous reaction? -> Investigate sign of overall reaction potential.

7.5 What is the value of the solubility product constant (Ksp) for AgCl?
AgCl  +  e-   -->  Ag  +  Cl-       E0 = 0.22 V
Ag    -->   Ag+ + e-                 E0 = -0.80 V
---------------------------------
AgCl --> Ag+2 + Cl-

Solution:
Step 1) Insert first and second half-reaction
AgCl + e- > Ag + Cl-     E0 = 0.22 V
Ag > Ag+ + e-                E0 = -0.80 V
and calculate equation and overall cell voltage.
Step 2) Insert n=1 (-n=-1) and E0  cell=-0.58 in equation: G = -n F E0 and calculate.
Step 3) Transfer result of G to equation: -G = R T ln K (remember to change sign of G) and calculate K in which represent the solubility product constant.

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