In a Daniell cell, $\mathrm{Zn} | \mathrm{Zn}^{2+}(aq) \,||\, \mathrm{…

Question

In a Daniell cell, $\mathrm{Zn} | \mathrm{Zn}^{2+}(aq) \,||\, \mathrm{Cu}^{2+}(aq) | \mathrm{Cu}$, the standard EMF is calculated as:

Accepted Answer

C. $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} - E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}}$

$E^\circ_	ext{cell} = E^\circ_	ext{cathode} - E^\circ_	ext{anode}$ where both standard electrode potentials are written as reduction potentials. Copper is the cathode (reduction $\mathrm{Cu}^{2+} + 2e^- 	o \mathrm{Cu}$) and zinc is the anode in a Daniell cell.

Answer

$E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} + E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}}$

Answer

$E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} - E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}$

Answer

$\dfrac{1}{2}\left(E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} + E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}}\right)$