For the cell reaction with the indicated conc | vedclass

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

Question

(A) The cell reaction is: $Al_{(s)} + 3Ag^{+}_{(aq)} 	o Al^{3+}_{(aq)} + 3Ag_{(s)}$The number of electrons transferred $(n)$ is $3$.The Nernst equati

Vedclass · Accepted Answer

$1.5496$

Vedclass · Answer

(A) The cell reaction is: $Al_{(s)} + 3Ag^{+}_{(aq)} 	o Al^{3+}_{(aq)} + 3Ag_{(s)}$The number of electrons transferred $(n)$ is $3$.The Nernst equation is given by: $E_{cell} = E^o_{cell} - \frac{0.06}{n} \log \frac{[Al^{3+}]}{[Ag^{+}]^3}$Substituting the given values: $1.50 = E^o_{cell} - \frac{0.06}{3} \log \frac{0.30}{(0.10)^3}$$1.50 = E^o_{cell} - 0.02 \log \frac{0.30}{0.001}$$1.50 = E^o_{cell} - 0.02 \log (300)$$1.50 = E^o_{cell} - 0.02 (\log 3 + \log 100)$$1.50 = E^o_{cell} - 0.02 (0.48 + 2)$$1.50 = E^o_{cell} - 0.02 (2.48)$$1.50 = E^o_{cell} - 0.0496$$E^o_{cell} = 1.50 + 0.0496 = 1.5496 \, V$

Similar Questions

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 ......... $V$.

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

If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?

The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module