Question 19.2: A galvanic cell consists of an Mg electrode in a 1.0 M Mg(NO...
A galvanic cell consists of an \text{Mg} electrode in a 1.0 M \text{Mg(NO}_3)_2 solution and a \text{Cd} electrode in a 1.0 M \text{Cd(NO}_3)_2 solution. Determine the overall cell reaction, and calculate the standard cell potential at 25°\text{C}.
Strategy Use the tabulated values of E^\circ to determine which electrode is the cathode and which is the anode, combine cathode and anode half-cell reactions to get the overall cell reaction, and use Equation 19.1 to calculate E^\circ_\text{cell}
Setup The half-cell reactions and their standard reduction potentials are
\text{Mg}^{2+} + 2e^– → \text{Mg} E^\circ = –2.37 \text{V}
\text{Cd}^{2+} + 2e^– → \text{Cd} E^\circ = –0.40 \text{V}
Because the Cd half-cell reaction has the greater (less negative) standard reduction potential, it will occur as the reduction. The Mg half-cell reaction will occur as the oxidation. Therefore, E^\circ_\text{cathode} = –0.40 \text{V} and E^\circ_\text{anode} = –2.37 \text{V}.
E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode} Equation 19.1
Adding the two half-cell reactions together gives the overall cell reaction:
\text{Mg} → \text{Mg}^{2+} + 2e^–
\underline{\text{Cd}^{2+} + 2e^– → \text{Cd} }
Overall: \text{Mg} + \text{Cd}^{2+} → \text{Mg}^{2+} + \text{Cd}
The standard cell potential is
E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}
= E^\circ_{\text{Cd}^{2+}/\text{Cd}} − E^\circ_{\text{Mg}^{2+}/\text{Mg}}
= (–0.40 \text{V}) − (–2.37 \text{V})
= 1.97 \text{V}