Given $E^{0}_{Fe^{2+} | Fe} = -0.44 \, V$ and $E^{0}_{Sn^{2+} | Sn} = -0.14 \, V$,calculate $E^{0}_{cell}$ for the reaction $Fe^{2+} + Sn \rightarrow Fe + Sn^{2+}$.

  • A
    $0.30$
  • B
    $-0.58$
  • C
    $0.58$
  • D
    $-0.30$

Explore More

Similar Questions

Calculate the standard cell potential for a cell having the following reaction: $2 Al_{(s)} + 3 Ni^{2+} \rightarrow 2 Al^{3+} + 3 Ni_{(s)}$ given that $E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$ and $E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$. (in $V$)

Using the standard reduction potentials of the electrodes $Li$,$Zn$,$Mg$,and $Ni$ as $-3.05 \ V$,$-0.76 \ V$,$-2.36 \ V$,and $-0.25 \ V$ respectively,identify the correct statement.

Given the standard electrode potentials,$K^{+}/K = -2.93 \, V$,$Ag^{+}/Ag = 0.80 \, V$,$Hg^{2+}/Hg = 0.79 \, V$,$Mg^{2+}/Mg = -2.37 \, V$,and $Cr^{3+}/Cr = -0.74 \, V$,arrange these metals in their increasing order of reducing power.

Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

Given $E_{Fe^{3+}|Fe}^0 = -0.036 \ V$ and $E_{Fe^{2+}|Fe}^0 = -0.439 \ V$,calculate $E^0_{cell}$ for the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$.

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