Calculate the cell potential for Ag_{(s)}|Ag^ | vedclass

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

Question

(A) The cell reaction is: $Ag_{(s)} + Ag^{+}_{(0.1 \ M)} \rightarrow Ag^{+}_{(0.01 \ M)} + Ag_{(s)}$.Using the Nernst equation at $298 \ K$:$E_{cell}

Vedclass · Accepted Answer

$0.0592$

Vedclass · Answer

(A) The cell reaction is: $Ag_{(s)} + Ag^{+}_{(0.1 \ M)} ightarrow Ag^{+}_{(0.01 \ M)} + Ag_{(s)}$.Using the Nernst equation at $298 \ K$:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ag^{+}]_{anode}}{[Ag^{+}]_{cathode}}$.Here,$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 \ V - 0.80 \ V = 0 \ V$.$n = 1$ (number of electrons transferred).$E_{cell} = 0 - \frac{0.0591}{1} \log \frac{0.01}{0.1}$.$E_{cell} = -0.0591 	imes \log(0.1) = -0.0591 	imes (-1) = 0.0591 \ V \approx 0.0592 \ V$.

Calculate the cell potential for $Ag_{(s)}|Ag^{+} \, (0.01 \ M)||Ag^{+} \, (0.1 \ M)|Ag_{(s)}$ at $298 \ K$. (in $V$)

Similar Questions

Give the $E_{cell}$ equation for the general reaction: $aA + bB \to cC + dD$.

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

For a $Mg|Mg^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag$ cell,the correct Nernst Equation is $:$

For the cell reaction,$3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$,$E^{\circ}_{cell}$ is $0.89 \ V$. Then $\Delta G^{\circ}$ for the reaction is

What will be the $emf$ for the given cell $Pt|H_2(P_1)|H^{+}_{(aq)}||H_2(P_2)|Pt$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module