For the cell Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 M) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The cell reaction is: $Cu_{(s)} + 2Ag^{+}_{(aq)} ightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$.The Nernst equation is: $E_{	ext{cell}} = E^{\circ}_{	e

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) The cell reaction is: $Cu_{(s)} + 2Ag^{+}_{(aq)} ightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$.The Nernst equation is: $E_{	ext{cell}} = E^{\circ}_{	ext{cell}} - \frac{0.059}{2} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.For the first cell: $0.3095 = E^{\circ}_{	ext{cell}} - \frac{0.059}{2} \log \frac{0.1}{(0.01)^2} = E^{\circ}_{	ext{cell}} - \frac{0.059}{2} \log(1000) = E^{\circ}_{	ext{cell}} - \frac{0.059}{2} 	imes 3$.$E^{\circ}_{	ext{cell}} = 0.3095 + 0.0885 = 0.398 \ V$.For the second cell: $E_{2} = 0.398 - \frac{0.059}{2} \log \frac{0.01}{(0.001)^2} = 0.398 - \frac{0.059}{2} \log(10000) = 0.398 - \frac{0.059}{2} 	imes 4 = 0.398 - 0.118 = 0.28 \ V$.$0.28 \ V = 28 	imes 10^{-2} \ V$.

Similar Questions

Which of the following will increase the voltage of the cell represented by the equation
$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$

For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)

What minimum decomposition potential is necessary to produce $Cl_2$ gas in the following reaction?
Given: $(\frac{2.303RT}{F} = 0.06)$
$Sn^{+2} (1 \ M) + 2Cl^{-} (2 \ M) \rightleftharpoons Sn_{(s)} + Cl_2 (1 \ atm)$
Given: $E^{o}_{Sn^{+2}/Sn} = -0.14 \ V$,$E^{o}_{Cl_2/Cl^{-}} = 1.4 \ V$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module