Calculate the E°cell in V for the following reaction
Pb + 2Ag+ → Pb2+ + 2Ag
The standard reduction potential for Pb2+/Pb & Ag+/Ag are −0.13 V and 0.80 V respectively
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.93 V
In a galvanic cell, oxidation takes place at the anode and reduction at the cathode. Using standard reduction potentials for both electrodes:
\( \displaystyle E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}} \)
In the given reaction, Pb is oxidised to \( \ce{Pb^{2+}} \) (anode) and \( \ce{Ag^{+}} \) is reduced to Ag (cathode):
\( \displaystyle \ce{Pb -> Pb^2+ + 2e^-}\quad(\text{anode}) \)
\( \displaystyle \ce{2Ag^+ + 2e^- -> 2Ag}\quad(\text{cathode}) \)
\( \displaystyle E^\circ_{\mathrm{cell}}=0.80-(-0.13) \)
\( \displaystyle E^\circ_{\mathrm{cell}}=0.93\ \mathrm{V} \)
The coefficient 2 in front of Ag is not used to multiply E°. The electrode potential is an intensive property; it does not depend on how many moles react.