For the cell Zn_{(s)} | Zn^{2+}_{(aq)} || M^{ | vedclass

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

Question

(A) The standard cell potential is given by $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.Given $E^o_{Zn^{2+}/Zn} = -0.76 \ V$,the anode is $Zn$.We calcu

Vedclass · Accepted Answer

$Ag^{+}/Ag$

Vedclass · Answer

(A) The standard cell potential is given by $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.Given $E^o_{Zn^{2+}/Zn} = -0.76 \ V$,the anode is $Zn$.We calculate $E^o_{cell}$ for each cathode:$1$. For $Au^{3+}/Au$ $(n=3)$: $E^o_{cell} = 1.40 - (-0.76) = 2.16 \ V$. Value per electron = $2.16 / 3 = 0.72 \ V$.$2$. For $Ag^{+}/Ag$ $(n=1)$: $E^o_{cell} = 0.80 - (-0.76) = 1.56 \ V$. Value per electron = $1.56 / 1 = 1.56 \ V$.$3$. For $Fe^{3+}/Fe^{2+}$ $(n=1)$: $E^o_{cell} = 0.77 - (-0.76) = 1.53 \ V$. Value per electron = $1.53 / 1 = 1.53 \ V$.$4$. For $Fe^{2+}/Fe$ $(n=2)$: $E^o_{cell} = -0.44 - (-0.76) = 0.32 \ V$. Value per electron = $0.32 / 2 = 0.16 \ V$.Comparing the values,$Ag^{+}/Ag$ gives the maximum value of $E^o_{cell}$ per electron transferred.

$M^{x+}_{(aq)} / M_{(s)}$	$Au^{3+}_{(aq)} / Au_{(s)}$	$Ag^{+}_{(aq)} / Ag_{(s)}$	$Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$	$Fe^{2+}_{(aq)} / Fe_{(s)}$
$E^o M^{x+} / M (V)$	$1.40$	$0.80$	$0.77$	$-0.44$

$ClO_4^{-}$	$E^{\circ} = 1.19 \ V$
$IO_4^{-}$	$E^{\circ} = 1.65 \ V$
$BrO_4^{-}$	$E^{\circ} = 1.74 \ V$

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 $V$)

When a rod of metal $A$ is dipped in an aqueous solution of metal $B$ (concentration of $B^{2+}$ ion being $1 \ M$) at $25 \ ^oC$,the standard electrode potentials are $E^o_{A^{2+}/A} = -0.76 \ V$ and $E^o_{B^{2+}/B} = +0.34 \ V$. What will happen?

Calculate $E_{cell}^o$ of the following galvanic cell at $298 \ K$:
$Ca_{(s)} | Ca^{2+}_{(aq)} || Fe^{2+}_{(aq)} | Fe_{(s)}$
Given: $E_{Ca^{2+}/Ca}^o = -2.87 \ V$; $E_{Fe/Fe^{2+}}^o = 0.41 \ V$

Reduction potential of ions are given below:

$ClO_4^{-}$ $E^{\circ} = 1.19 \ V$

$IO_4^{-}$ $E^{\circ} = 1.65 \ V$

$BrO_4^{-}$ $E^{\circ} = 1.74 \ V$

The correct order of their oxidising power is:

Identify the representation of the standard hydrogen electrode.

Vedclass Test Series

Exam Paper Generator

Online Exam Module