Calculate the potential (in volts) for the voltaic (or galvanic) cell indicated at $25^{\circ} \math

Question

Calculate the potential (in volts) for the voltaic (or galvanic) cell indicated at $25^{\circ} \mathrm{C}$.\ $\mathrm{Ga} / \mathrm{Ga}^{3+}\left(10^{-6} \mathrm{M}ight)| | \mathrm{Ag}^{+}\left(10^{-4} \mathrm{M}ight) / \mathrm{Ag}$\ ( $\mathrm{E}_{0} \mathrm{Ga}^{3+} / \mathrm{Ga}=-0.56 \mathrm{~V} \mathrm{Ag}+/ \mathrm{Ag}=0.8 \mathrm{~V}$ )\

Professor PC Thomas Classes · Answer

$\mathrm{Ga}+3 \mathrm{Ag}^{+} ightarrow \mathrm{Ga}^{3+}+3 \mathrm{Ag}$\ $\mathrm{E}^{0}=\mathrm{E}_{	ext {cathode }}^{0}-\mathrm{E}_{	ext {anode }}^{0}=0.8-(-0.56)=1.36 \mathrm{~V}$.\ $\mathrm{E}=\mathrm{E}^{0}-\frac{0.0591}{\mathrm{n}} \log \frac{[\mathrm{P}]}{[\mathrm{R}]}=1.36-\frac{0.0591}{3} \log \frac{10^{-6}}{\left(10^{-4}ight)^{3}}$\ $=1.36-\frac{0.0591}{3} 	imes 6=1.36-0.1182=1.2418 \mathrm{~V}$\