If E^{circ}(Al_{(aq)}^{+3} mid Al_{(s)}) = -1 | vedclass

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

Question

(C) The given half-cell reaction is the oxidation of aluminum: $Al_{(s)} \rightarrow Al_{(aq)}^{+3} + 3e^-$. The standard reduction potential is $E^{\

Vedclass · Accepted Answer

$+1.679 \ V$

Vedclass · Answer

(C) The given half-cell reaction is the oxidation of aluminum: $Al_{(s)} ightarrow Al_{(aq)}^{+3} + 3e^-$. The standard reduction potential is $E^{\circ}(Al^{+3} \mid Al) = -1.66 \ V$,so the standard oxidation potential is $E^{\circ}_{ox} = +1.66 \ V$. Using the Nernst equation for the oxidation potential: $E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log[Al^{+3}]$. Here,$n = 3$ and $[Al^{+3}] = 0.1 \ M$. $E_{ox} = 1.66 - \frac{0.0591}{3} \log(0.1)$. $E_{ox} = 1.66 - 0.0197 	imes (-1)$. $E_{ox} = 1.66 + 0.0197 = 1.6797 \ V \approx +1.679 \ V$.

If $E^{\circ}(Al_{(aq)}^{+3} \mid Al_{(s)}) = -1.66 \ V$. What is the potential of $Al_{(s)} \rightarrow Al_{(aq)}^{+3}(0.1 \ M) + 3e^-$ at $298 \ K$?

Similar Questions

Which of the following is the correct representation of the Nernst equation for the reduction reaction $M^{n+} + ne^- \rightarrow M$?

At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

Calculate the equilibrium constant of the reaction,$Cu_{(s)} + 2 Ag^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$,given that for the reaction $E^{\circ}_{cell} = 0.46 \ V$.

$E_{1}$,$E_{2}$ and $E_{3}$ are the emfs of the following three galvanic cells respectively.

What is the potential of a cell containing two hydrogen electrodes,the negative one in contact with $10^{-8} \ M \ H^{+}$ and the positive one in contact with $0.025 \ M \ H^{+}$? (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module