What will be the electrode potential of a Cu | vedclass

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

Question

(B) The reduction half-reaction is: $Cu^{2+} + 2e^{-} \rightarrow Cu(s)$ Using the Nernst equation: $E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0

Vedclass · Accepted Answer

$0.293$

Vedclass · Answer

(B) The reduction half-reaction is: $Cu^{2+} + 2e^{-} ightarrow Cu(s)$ Using the Nernst equation: $E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$ Given $E^{\circ} = 0.34 \ V$,$n = 2$,and $[Cu^{2+}] = 0.025 \ M$: $E = 0.34 - \frac{0.0591}{2} \log \frac{1}{0.025}$ $E = 0.34 - 0.02955 	imes \log(40)$ Since $\log(40) = \log(4 	imes 10) = 0.602 + 1 = 1.602$: $E = 0.34 - 0.02955 	imes 1.602 \approx 0.34 - 0.0473 = 0.2927 \ V$ Rounding to three decimal places,we get $0.293 \ V$.

What will be the electrode potential of a $Cu$ electrode dipped in $0.025 \ M \ CuSO_4$ solution at $298 \ K$,given that the standard reduction potential of $Cu$ is $0.34 \ V$ (in $V$)?

Similar Questions

The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

Calculate the $E^o_{cell}$ for the following reaction: $Cu^{+2}_{(aq)} + Sn^{+2}_{(aq)} \rightarrow Cu_{(s)} + Sn^{+4}_{(aq)}$,given that the equilibrium constant $K_c = 10^6$. (in $V$)

For the reaction; $2A_{(s)} + B_{(aq)}^{2+} \to 2A_{(aq)}^{+} + B_{(s)}$,the Nernst equation for the $EMF$ of the cell is:

Which will increase the voltage of the cell $Sn_{(s)} + 2Ag^{+}_{(aq)} \to Sn^{2+}_{(aq)} + 2Ag_{(s)}$?

Which Nernst equation is correct for the following cell? $Al_{(s)}|Al_{(aq)}^{3+} || Zn_{(aq)}^{2+}| Zn_{(s)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module