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 :

Similar Questions

Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)

The standard $e.m.f.$ of a cell,involving one electron change is found to be $0.591 \ V$ at $25^{\circ} C$. The equilibrium constant of the reaction is :
$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$

Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

What must be the concentration of $Ag^{+}$ in an aqueous solution containing $Cu^{2+} = 1.0 \ M$ so that both the metals can be deposited on the cathode simultaneously? Given that $E^0_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^0_{Ag^{+}/Ag} = 0.812 \ V$ at $T = 298 \ K$.

Calculate the equilibrium constant of the reaction,$Cu_{(s)} + 2 Ag^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$,given that for the reaction $E^{\circ}_{cell} = 0.46 \ V$.

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