Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Step by step solution

01

Identify the half reactions

It is given that copper electrode is the anode and zinc penny is the cathode. In an electrochemical cell, oxidation happens at the anode and reduction at the cathode. Thus, the two relevant reactions are: \(Cu(s) \rightarrow Cu^{2+} + 2e^-\) at the anode and \(Zn^{2+} + 2e^- \rightarrow Zn(s)\) at the cathode. Note that the reactions are reversed from the table because the copper reaction is actually an oxidation.

02

Calculate the voltage difference

The total voltage of the cell (E°cell) is equal to the difference between the reduction potentials of the cathode and the anode. So we have: E°cell = E°cathode – E°anode = (-0.76 V) - (+0.34 V) = -1.10 V.

03

Correct the sign

We've found that the cell potential is -1.10 V. But when talking about required voltages for electrochemical cells to function we always give positive values, because you can think of this as the amount of 'push' needed to make the cell run. So we correct the sign to get +1.10 V as our answer.

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