Calculate E_{cell} of the reaction Mg_{(s)} + | vedclass

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(B) Given reaction: $Mg_{(s)} + 2Ag^{+}_{(aq)} 	o Mg^{2+}_{(aq)} + 2Ag_{(s)}$Here,the number of electrons transferred $n = 2$.Using the Nernst equati

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$+2.96$

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(B) Given reaction: $Mg_{(s)} + 2Ag^{+}_{(aq)} 	o Mg^{2+}_{(aq)} + 2Ag_{(s)}$Here,the number of electrons transferred $n = 2$.Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}$$E_{cell} = 3.17 - \frac{0.0591}{2} \log \frac{0.100}{(0.0001)^2}$$E_{cell} = 3.17 - 0.02955 \log \frac{10^{-1}}{10^{-8}}$$E_{cell} = 3.17 - 0.02955 \log 10^7$$E_{cell} = 3.17 - 0.02955 	imes 7$$E_{cell} = 3.17 - 0.20685$$E_{cell} \approx 2.96 \ V$

Calculate $E_{cell}$ of the reaction $Mg_{(s)} + 2Ag^{+}_{(aq)} (0.0001 \ M) \to Mg^{2+}_{(aq)} (0.100 \ M) + 2Ag_{(s)}$ in $V$. If $E^o_{cell} = 3.17 \ V$.

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