For a cell,Cu_{(s)} mid Cu^{2+}(0.001,M) midm | vedclass

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

Question

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

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(C) The cell reaction is: $Cu_{(s)} + 2Ag^{+}_{(aq)} ightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{	ext{cell}} = E^{\Theta}_{	ext{cell}} - \frac{0.06}{n} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.Here,$n = 2$,$[Cu^{2+}] = 10^{-3} \, M$,and $[Ag^{+}] = 10^{-2} \, M$.$0.43 = E^{\Theta}_{	ext{cell}} - \frac{0.06}{2} \log \frac{10^{-3}}{(10^{-2})^2}$.$0.43 = E^{\Theta}_{	ext{cell}} - 0.03 \log \frac{10^{-3}}{10^{-4}} = E^{\Theta}_{	ext{cell}} - 0.03 \log(10) = E^{\Theta}_{	ext{cell}} - 0.03$.$E^{\Theta}_{	ext{cell}} = 0.43 + 0.03 = 0.46 \, V$.We know $E^{\Theta}_{	ext{cell}} = E^{\Theta}_{	ext{cathode}} - E^{\Theta}_{	ext{anode}} = E^{\Theta}_{Ag^{+}/Ag} - E^{\Theta}_{Cu^{2+}/Cu}$.$0.46 = 0.80 - E^{\Theta}_{Cu^{2+}/Cu}$.$E^{\Theta}_{Cu^{2+}/Cu} = 0.80 - 0.46 = 0.34 \, V$.Thus,$E^{\Theta}_{Cu^{2+}/Cu} = 34 	imes 10^{-2} \, V$.

Similar Questions

In which of the following conditions will the reduction potential of a hydrogen half-cell be negative?

For the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when:

If the $E^{\circ}_{cell}$ for a given reaction has a negative value,which of the following gives the correct relationships for the values of $\Delta G^{\circ}$ and $K_{eq}$ ?

Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

What is the reduction potential of a silver wire dipped in a $0.1 \ M \ AgNO_3$ solution at $25^\circ C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module