What happens to the emf of the cell Zn_{(s)} | vedclass

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

Question

(B) The cell reaction is: $Zn_{(s)} + 2Ag^{+1}_{(aq)} \rightarrow Zn^{+2}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \fra

Vedclass · Accepted Answer

decrease by $0.0592 \ V$

Vedclass · Answer

(B) The cell reaction is: $Zn_{(s)} + 2Ag^{+1}_{(aq)} ightarrow Zn^{+2}_{(aq)} + 2Ag_{(s)}$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q$.Here,$n = 2$ and $Q = \frac{[Zn^{+2}]}{[Ag^{+1}]^2}$.Initially,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 1 \ M$,so $Q_1 = \frac{1}{1^2} = 1$. Thus,$E_1 = E^0_{cell} - \frac{0.0592}{2} \log(1) = E^0_{cell}$.Finally,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 0.1 \ M$,so $Q_2 = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.$E_2 = E^0_{cell} - \frac{0.0592}{2} \log(100) = E^0_{cell} - 0.0296 	imes 2 = E^0_{cell} - 0.0592 \ V$.The change in $emf$ is $E_2 - E_1 = -0.0592 \ V$,which means the $emf$ decreases by $0.0592 \ V$.

What happens to the $emf$ of the cell $Zn_{(s)} | Zn^{+2} (1 \ M) || Ag^{+1} (1 \ M) | Ag_{(s)}$ if the concentration of $Ag^{+1}$ decreases to $0.1 \ M$?

Similar Questions

The reduction potential of a hydrogen electrode at $pH = 1$ is ........ $V$.

At a temperature of $298 \ K$, the $emf$ of the following electrochemical cell: $Ag_{(s)} | Ag^{+}(0.1 \ M) || Zn^{2+}(0.1 \ M) | Zn_{(s)}$ will be (Given, $E^{\circ}_{cell} = -1.562 \ V$) (in $V$)

What is the standard free energy change for the cell,having the following cell reaction (in $kJ$)?
$2 Ag_{(aq)}^{+} + Cd_{(s)} \longrightarrow 2 Ag_{(s)} + Cd_{(aq)}^{2+}, E^{\circ}_{cell} = 1.20 \ V$

For the cell reaction,$Cu | Cu^{2+}(0.1 \ M) || Cu^{2+}(1.0 \ M) | Cu$,the emf of the cell at $25^{\circ}C$ is given that $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$. (in $V$)

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module