Consider the electrochemical cell shown in th | vedclass

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

Question

(C) For a concentration cell, the cell reaction is $M^{+}(c_{cathode}) \rightarrow M^{+}(c_{anode})$.Using the Nernst equation: $E_{cell} = E^{\circ}_

Vedclass · Accepted Answer

If $c_{1}$ is present at the cathode, then $c_{1} > c_{2}$

Vedclass · Answer

(C) For a concentration cell, the cell reaction is $M^{+}(c_{cathode}) \rightarrow M^{+}(c_{anode})$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[M^{+}]_{anode}}{[M^{+}]_{cathode}}$.Since $E^{\circ}_{cell} = 0$ for a concentration cell, $E_{cell} = -0.0591 \log \frac{c_{anode}}{c_{cathode}} = 0.0591 \log \frac{c_{cathode}}{c_{anode}}$.For $E_{cell} > 0$, we must have $\log \frac{c_{cathode}}{c_{anode}} > 0$, which implies $c_{cathode} > c_{anode}$.Case $I$: If $c_{1}$ is at the cathode, then $c_{1} > c_{2}$ (Option $C$ is correct).Case $II$: If $c_{1}$ is at the anode, then $c_{2} > c_{1}$ (Option $D$ is incorrect).

Similar Questions

What is the oxidation potential of $0.05 \, M \, H_2SO_4$ in volts?

Fill in the blanks :
$1.$ The ratio of concentration of products to concentration of reactants is ........
$2.$ $\ln(\log(x)) =$ ............
$3.$ At equilibrium,between $E_{cell}$ and $E_{cell}^{o}$,......... will be zero.

Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

For the cell reaction,$A_{(s)} + B^{2+}_{(aq)} \rightarrow A^{2+}_{(aq)} + B_{(s)}$,if the equilibrium constant of the reaction is $10^4$ at $298 \ K$,what is the standard $EMF$ of the cell (in $V$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module