| Given the following half reactions: |
| \[{{A}^{2+}}+{{e}^{-}}\to {{A}^{+}},{{E}^{o}}=+1.70V\] |
| \[{{B}^{2+}}+2{{e}^{-}}\to {{B}^{+}},{{E}^{o}}=+0.60V\] |
| \[{{C}^{3+}}+{{e}^{-}}\to {{C}^{2+}},{{E}^{o}}=+0.90V\] |
| The only reaction that could be used in titration would be |
A) \[{{A}^{+}}\] with \[{{B}^{4+}}\]
B) \[{{A}^{+}}\] with \[{{C}^{3+}}\]
C) \[{{A}^{2+}}\] with \[{{C}^{2+}}\]
D) \[{{C}^{2+}}\] with \[{{B}^{4+}}\]
Correct Answer: C
Solution :
[c] \[\begin{align} & \,\,\,\,\,{{A}^{2}}+{{e}^{-}}\to {{A}^{+}},E=1.7V \\ & \,\,\,\,\,\,\,\,\,\,\,\,{{C}^{2+}}\to {{C}^{3+}}+{{e}^{-}},{{E}^{\circ }}=-0.9V \\ & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ & {{A}^{2+}}+{{C}^{2+}}\to {{A}^{+}}+{{C}^{3+}},{{E}^{\circ }}=0.8V>0, \\ \end{align}\]spontaneous.
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