Cell Potential and Free Energy - AP Chemistry

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Question

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

Answer

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half reaction becomes negative. $\small Cu$ half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

Compare your answer with the correct one above

Question

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

Answer

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half reaction becomes negative. $\small Cu$ half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

Compare your answer with the correct one above

Question

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

Answer

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half reaction becomes negative. $\small Cu$ half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

Compare your answer with the correct one above

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