E_{cell}^{0} of the reaction Mg_{(s)} + 2 Ag_ | vedclass

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Question

(C) The cell reaction is $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$.Here,$n = 2$.The Nernst equation is $E_{cell} = E_{cell}

Vedclass · Accepted Answer

$2.993 \ V, Mg | Mg_{(0.01 \ M)}^{2+} || Ag_{(0.0001 \ M)}^{+} | Ag$

Vedclass · Answer

(C) The cell reaction is $Mg_{(s)} + 2 Ag_{(aq)}^{+} ightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$.Here,$n = 2$.The Nernst equation is $E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}$.Given $E_{cell}^{0} = 3.17 \ V$,$[Mg^{2+}] = 0.01 \ M$,and $[Ag^{+}] = 0.0001 \ M$.$E_{cell} = 3.17 - \frac{0.0591}{2} \log \frac{0.01}{(0.0001)^2}$.$E_{cell} = 3.17 - 0.02955 \log \frac{10^{-2}}{10^{-8}} = 3.17 - 0.02955 \log(10^6)$.$E_{cell} = 3.17 - 0.02955 	imes 6 = 3.17 - 0.1773 = 2.9927 \ V \approx 2.993 \ V$.The cell notation is written as $	ext{Anode} | 	ext{Anode electrolyte} || 	ext{Cathode electrolyte} | 	ext{Cathode}$.Thus,$Mg | Mg_{(0.01 \ M)}^{2+} || Ag_{(0.0001 \ M)}^{+} | Ag$.

$E_{cell}^{0}$ of the reaction $Mg_{(s)} + 2 Ag_{(0.0001 \ M)}^{+} \rightleftharpoons Mg_{(0.01 \ M)}^{2+} + 2 Ag_{(s)}$ is $3.17 \ V$. The $E_{cell}$ of the reaction and its cell notation respectively are :

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