Electrochemistry Question 2

A galvanic cell is constructed using two half-cells based on the two half-reactions represented below.

\ce{Zn^{2+}(aq) + 2e^- -> Zn(s)} \hskip{2em} E^\circ = \pu{-0.76 V}

\ce{Fe^{3+}(aq) + e^- -> Fe^{2+}(aq)} \hskip{2em} E^\circ = \pu{0.77 V}

What is the standard cell potential for the galvanic cell?

-0.01

0.01

1.53

2.31

Approach

To obtain the standard cell potential, we add the cell potential of each half-cell reaction.

The reduction of $\ce{Fe^{3+}}$ is more likely since the cell potential is greater. Therefore, zinc will be oxidized. We can reverse the equation and change the sign of the cell potential.

Reduction (Cathode):

\ce{Fe^{3+}(aq) + e^- -> Fe^{2+}(aq)} \hskip{2em} E^\circ = \pu{0.77 V}

Oxidation (Anode):

\ce{Zn^{2+}(aq) + 2e^- -> Zn(s)} \hskip{2em} E^\circ = \pu{-0.76 V}

\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-} \hskip{2em} E^\circ = \pu{0.76 V}

Adding the half-cell potentials:

0.77 + 0.76 = \boxed{1.53}