To find the standard potential of M^{3+}/M el | vedclass

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

Question

(B) The cell reaction is: $M(s) + 3Ag^{+}(aq) 	o M^{3+}(aq) + 3Ag(s)$.The number of electrons transferred is $n = 3$.Using the Nernst equation: $E_{c

Vedclass · Accepted Answer

$0.32$

Vedclass · Answer

(B) The cell reaction is: $M(s) + 3Ag^{+}(aq) 	o M^{3+}(aq) + 3Ag(s)$.The number of electrons transferred is $n = 3$.Using the Nernst equation: $E_{cell} = E^{o}_{cell} - \frac{0.0591}{n} \log \frac{[M^{3+}]}{[Ag^{+}]^{3}}$.Substituting the given values: $0.421 = E^{o}_{cell} - \frac{0.0591}{3} \log \frac{0.001}{(0.01)^{3}}$.$0.421 = E^{o}_{cell} - 0.0197 \log \frac{10^{-3}}{10^{-6}} = E^{o}_{cell} - 0.0197 \log(10^{3})$.$0.421 = E^{o}_{cell} - 0.0197 	imes 3 = E^{o}_{cell} - 0.0591$.$E^{o}_{cell} = 0.421 + 0.0591 = 0.4801 \ V \approx 0.48 \ V$.Since $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = E^{o}_{Ag^{+}/Ag} - E^{o}_{M^{3+}/M}$.$0.48 = 0.80 - E^{o}_{M^{3+}/M}$.$E^{o}_{M^{3+}/M} = 0.80 - 0.48 = 0.32 \ V$.

Similar Questions

For a reaction,$A_{(s)} + 2B_{(aq)}^{+} \rightleftharpoons A_{(aq)}^{2+} + 2B_{(s)}$,$K_{c}$ is $10^{12}$ at $25^{\circ} C$. The $E_{Cell}^{\circ}$ of the corresponding cell is $(F = 96500 \ C \ mol^{-1})$ (in $V$)

Given $E^o _{Cr^{3+} / Cr} = -0.72 \, V$ and $E^o _{Fe^{2+} / Fe} = -0.42 \, V$. The potential for the cell $Cr \, | \, Cr^{3+}_{(0.1 \, M)} \, || \, Fe^{2+}_{(0.01 \, M)} \, | \, Fe$ is ......... $V$.

What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^{\circ}_{Cu^{+2}/Cu} = 0.34 \ V$.

Calculate the $E_{cell}$ for $Zn_{(s)} | Zn^{2+}_{(0.1 \ M)} || Cr^{3+}_{(0.1 \ M)} | Cr_{(s)}$ at $25^{\circ} C$ if $E^{\circ}_{cell}$ is $0.02 \ V$. (in $V$)

Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module