One half cell in a voltaic cell is constructe | vedclass

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

Question

(B) $1$. The cell reaction is: $Zn(s) + 2Ag^+(aq) 	o Zn^{2+}(aq) + 2Ag(s)$.$2$. Calculate the standard cell potential: $E^\circ_{cell} = E^\circ_{cat

Vedclass · Accepted Answer

$\frac{4}{5.9}$

Vedclass · Answer

(B) $1$. The cell reaction is: $Zn(s) + 2Ag^+(aq) 	o Zn^{2+}(aq) + 2Ag(s)$.$2$. Calculate the standard cell potential: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 	ext{ V} - (-0.76 	ext{ V}) = 1.56 	ext{ V}$.$3$. Apply the Nernst equation at $298 	ext{ K}$: $E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$.$4$. Here, $n = 2$, $[Zn^{2+}] = 1 	ext{ M}$, and $[Ag^+] = x$. Substituting the values: $1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{x^2}$.$5$. Simplify the equation: $1.60 = 1.56 - 0.0295 	imes (-2 \log x) = 1.56 + 0.059 \log x$.$6$. Solve for $\log x$: $0.04 = 0.059 \log x$, which gives $\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$.

Similar Questions

$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

The equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is:

The standard cell potential for $Zn \mid Zn^{2+} \parallel Cu^{2+} \mid Cu$ is $1.10 \ V$. When the cell is completely discharged,$\log ([Zn^{2+}] / [Cu^{2+}])$ is closest to $.....$

$E_{cell}^{0}$ of the reaction $Mg_{(s)} + 2 Ag_{(0.0001 \ M)}^{+} \rightleftharpoons Mg_{(0.01 \ M)}^{2+} + 2 Ag_{(s)}$ is $3.17 \ V$. The $E_{cell}$ of the reaction and its cell notation respectively are :

Which of the following statements is correct regarding the $emf$ of the cell for the cell reaction,$Cd_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Cd^{2+}_{(aq)} + Cu_{(s)}$,if the concentration of $Cd^{2+}$ is $10$ times greater than the concentration of $Cu^{2+}_{(aq)}$ at $298 \ K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module