The electromotive force for the following cel | vedclass

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

Question

(C) For the given concentration cell,the cell reaction is: $Cu_{(s)} + Cu^{+2} (0.04 \ M) \rightarrow Cu^{+2} (0.01 \ M) + Cu_{(s)}$.Applying the Nern

Vedclass · Accepted Answer

$ - \frac{RT}{F} \ln (0.5) $

Vedclass · Answer

(C) For the given concentration cell,the cell reaction is: $Cu_{(s)} + Cu^{+2} (0.04 \ M) ightarrow Cu^{+2} (0.01 \ M) + Cu_{(s)}$.Applying the Nernst equation: $E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q$.Here,$E^0_{cell} = 0$ for a concentration cell,$n = 2$,and $Q = \frac{[Cu^{+2}]_{anode}}{[Cu^{+2}]_{cathode}} = \frac{0.01}{0.04} = 0.25$.Substituting these values: $E_{cell} = 0 - \frac{RT}{2F} \ln (0.25)$.Since $0.25 = (0.5)^2$,we have $\ln (0.25) = 2 \ln (0.5)$.Therefore,$E_{cell} = - \frac{RT}{2F} 	imes 2 \ln (0.5) = - \frac{RT}{F} \ln (0.5)$.

The electromotive force for the following cell $Cu_{(s)} | Cu^{+2} (0.01 \ M) || Cu^{+2} (0.04 \ M) | Cu_{(s)}$ is

Similar Questions

Calculate the cell potential for a $Cu$ plate kept in $0.2 \ M$ $CuSO_4$ solution. Given: $E_{Cu^{2+} \mid Cu}^o = 0.34 \ V$. (in $V$)

$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

What is the potential of a hydrogen electrode in contact with a solution whose $pH$ is $1$ (in $V$)?

The e.m.f. of the cell $Ag | Ag^{+}(0.1 \ M) || Ag^{+}(1 \ M) | Ag$ at $298 \ K$ is ........... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module