(C) The cell reaction is: $2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$Reduction half-reaction: $2Fe^{3+} + 2e^- \to 2Fe^{2+}$; $E^o_{red} = 0.771 \, V$Oxidati

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

(C) The cell reaction is: $2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$Reduction half-reaction: $2Fe^{3+} + 2e^- \to 2Fe^{2+}$; $E^o_{red} = 0.771 \, V$Oxidation half-reaction: $2I^- \to I_2 + 2e^-$; $E^o_{ox} = -E^o_{red} = -0.536 \, V$$E^o_{cell} = E^o_{red} + E^o_{ox} = 0.771 \, V + (-0.536 \, V) = 0.235 \, V$

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Medium

The electrode potentials are given as follows:
$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$
$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$
For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is:

A
$(2 \times 0.771 - 0.536) = 1.006 \, V$
B
$(0.771 - 0.5 \times 0.536) = 0.503 \, V$
C
$0.771 - 0.536 = 0.235 \, V$
D
$0.536 - 0.771 = -0.235 \, V$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Electrode potential and ECell

For the cell reaction $Mg_{(s)} + Zn^{2+}_{(aq)} (1M) \rightarrow Zn_{(s)} + Mg^{2+}_{(aq)} (1M)$,the $emf$ is $1.60 \ V$. The $E$ of the cell is ........ $V$.

Medium

View Solution

The standard electrode potential for the two electrodes $A^{+}/A$ and $B^{+}/B$ are respectively $0.5 \ V$ and $0.75 \ V$. The $EMF$ of the given cell $A | A^{+}(a = 1) || B^{+}(a = 1) | B$ will be $.......... \ V$. (in $.25$)

Medium

View Solution

The standard electrode potential for a $Daniell$ cell is $1.1 \ V$. Calculate the standard Gibbs energy for the reaction: $[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}]$

Medium

View Solution

Which of the following expressions is correct for the standard Gibbs free energy change?

Easy

View Solution

Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be:

DifficultJEE MAIN 2013

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The electrode potentials are given as follows:$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is:

Similar Questions

For the cell reaction $Mg_{(s)} + Zn^{2+}_{(aq)} (1M) \rightarrow Zn_{(s)} + Mg^{2+}_{(aq)} (1M)$,the $emf$ is $1.60 \ V$. The $E$ of the cell is ........ $V$.

The standard electrode potential for the two electrodes $A^{+}/A$ and $B^{+}/B$ are respectively $0.5 \ V$ and $0.75 \ V$. The $EMF$ of the given cell $A | A^{+}(a = 1) || B^{+}(a = 1) | B$ will be $.......... \ V$. (in $.25$)

The standard electrode potential for a $Daniell$ cell is $1.1 \ V$. Calculate the standard Gibbs energy for the reaction: $[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}]$

Which of the following expressions is correct for the standard Gibbs free energy change?

Electrode potentials $(E^o)$ are given below:$Cu^{+}/Cu = +0.52 \ V$$Fe^{3+}/Fe^{2+} = +0.77 \ V$$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$$Ag^{+}/Ag = +0.88 \ V$Based on the above potentials,the strongest oxidizing agent will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The electrode potentials are given as follows:
$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$
$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$
For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is:

Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be: