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Question

(B) The cell reaction is: $Mg_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Mg^{2+}_{(aq)} + 2Ag_{(s)}$.Standard cell potential $E^{\circ}_{cell} = E^{\circ}_{ca

Vedclass · Accepted Answer

$3.07$

Vedclass · Answer

(B) The cell reaction is: $Mg_{(s)} + 2Ag^{+}_{(aq)} ightarrow Mg^{2+}_{(aq)} + 2Ag_{(s)}$.Standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 \ V - (-2.37 \ V) = 3.17 \ V$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}$.Here,$n = 2$,$[Mg^{2+}] = 0.18 \ M$,and $[Ag^{+}] = 0.01 \ M$.$E_{cell} = 3.17 - \frac{0.0591}{2} \log \frac{0.18}{(0.01)^2} = 3.17 - 0.02955 \log \frac{0.18}{0.0001} = 3.17 - 0.02955 \log(1800)$.$E_{cell} = 3.17 - 0.02955 	imes 3.255 = 3.17 - 0.096 = 3.074 \ V$.

Calculate the cell potential for the reaction $Mg_{(s)} \mid Mg^{2+}(0.18 \ M) \parallel Ag^{+}(0.01 \ M) \mid Ag_{(s)}$. Given standard electrode potentials are $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$. (in $V$)

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