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Question

(B) The cell reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$.Here,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to $Fe$ (Cathode)

Vedclass · Accepted Answer

$0.35$

Vedclass · Answer

(B) The cell reaction is $Fe^{2+} + Zn ightarrow Zn^{2+} + Fe$.Here,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to $Fe$ (Cathode).Given oxidation potentials are:$E^{0}_{Zn|Zn^{2+}} = 0.76 \ V$$E^{0}_{Fe|Fe^{2+}} = 0.41 \ V$Standard cell potential $E^{0}_{cell} = E^{0}_{oxi}(	ext{anode}) + E^{0}_{red}(	ext{cathode})$.Since $E^{0}_{red}(	ext{cathode}) = -E^{0}_{oxi}(	ext{cathode})$,$E^{0}_{cell} = E^{0}_{oxi}(Zn|Zn^{2+}) - E^{0}_{oxi}(Fe|Fe^{2+})$$E^{0}_{cell} = 0.76 \ V - 0.41 \ V = 0.35 \ V$.

Calculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)

Similar Questions

For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .
$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$
$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)

The correct order of reduction potentials of the following pairs is
$A.$ $Cl_2 / Cl^{-}$
$B.$ $I_2 / I^{-}$
$C.$ $Ag^{+} / Ag$
$D.$ $Na^{+} / Na$
$E.$ $Li^{+} / Li$
Choose the correct answer from the options given below.

Which of the following reactions will be spontaneous in an electrochemical cell constructed from the given standard electrode potentials: $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$?

Given that $E^o_{K^{+}/K} = -2.93 \ V$,$E^o_{Fe^{2+}/Fe} = -0.44 \ V$,$E^o_{Zn^{2+}/Zn} = -0.76 \ V$,and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$. Based on this data,which of the following is the strongest reducing agent?

What does the negative sign in the expression $E_{Zn^{2+}|Zn}^o = -0.76 \ V$ mean?

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Calculate the cell potential in volts for the cell reaction given below:$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$Given:$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)