Question
Following cell is set up between copper and silver electrodes.$\text{Cu}|\text{Cu}^{2+}\text{(aq)}\|\text{Ag}^+\text{(aq)}|\text{Ag}$
If its two half cells work under standard conditions, calculate the emf of the cell.
$\Big[\text{Given E}^\circ_{\frac{\text{Cu}^{2+}}{\text{Cu}}}=+0.34\text{V},\text{E}^\circ_{\frac{\text{Ag}^+}{\text{Ag}}}=+0.80\text{V}\Big]$

Answer

$\text{E}^{\circ}_\text{cell}=\text{E}^\circ_\text{cathode}-\text{E}^\circ_\text{anode}$
$=\text{E}^\circ_{\frac{\text{Ag}^+}{\text{Ag}}}-\text{E}^\circ_{\frac{\text{Cu}^{2+}}{\text{Cu}}}=+0.80-(+0.34)=0.46\text{V}$

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