What is the procedure to determine the cell potential?
(N/A) Selection of reference electrode: The potential of an individual half-cell cannot be measured directly. We can only measure the difference between the potentials of two half-cells,which gives the $EMF$ of the cell.
$(b)$ Use of Standard Hydrogen Electrode $(SHE)$: To determine the potential of a single half-cell,it is coupled with a reference electrode whose potential is known. By convention,the Standard Hydrogen Electrode $(SHE)$ is used as the reference,and its potential is arbitrarily assigned a value of $0.00 \ V$ at all temperatures.
$(c)$ Calculation: When the half-cell is connected to the $SHE$,the measured $EMF$ of the resulting cell corresponds to the potential of that half-cell. If the half-cell acts as the cathode,its potential is positive; if it acts as the anode,its potential is negative relative to the $SHE$.
$(b)$ Use of Standard Hydrogen Electrode $(SHE)$: To determine the potential of a single half-cell,it is coupled with a reference electrode whose potential is known. By convention,the Standard Hydrogen Electrode $(SHE)$ is used as the reference,and its potential is arbitrarily assigned a value of $0.00 \ V$ at all temperatures.
$(c)$ Calculation: When the half-cell is connected to the $SHE$,the measured $EMF$ of the resulting cell corresponds to the potential of that half-cell. If the half-cell acts as the cathode,its potential is positive; if it acts as the anode,its potential is negative relative to the $SHE$.
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Similar Questions
The oxidation potentials of the following half-cell reactions are given:
$Zn \to Zn^{2+} + 2e^-; E^o = 0.76 \, V$
$Fe \to Fe^{2+} + 2e^-; E^o = 0.44 \, V$
What will be the $EMF$ of the cell whose cell reaction is:
$Fe^{2+}_{(aq)} + Zn \to Zn^{2+}_{(aq)} + Fe$
$Zn \to Zn^{2+} + 2e^-; E^o = 0.76 \, V$
$Fe \to Fe^{2+} + 2e^-; E^o = 0.44 \, V$
What will be the $EMF$ of the cell whose cell reaction is:
$Fe^{2+}_{(aq)} + Zn \to Zn^{2+}_{(aq)} + Fe$
Medium
View Solution$FeO_4^{2-}$ $\xrightarrow{2.2 \ V} Fe^{3+}$ $\xrightarrow{0.70 \ V} Fe^{2+}$ $\xrightarrow{-0.45 \ V} Fe^0$
$E_{FeO_4^{2-} / Fe^{2+}}^{\theta}$ is $x \times 10^{-3} \ V$. The value of $x$ is $.........$.
$E_{FeO_4^{2-} / Fe^{2+}}^{\theta}$ is $x \times 10^{-3} \ V$. The value of $x$ is $.........$.
DifficultJEE MAIN 2023
View SolutionIf the half-cell reaction $A + e^- \to A^-$ has a large negative reduction potential,it follows that
Medium
View Solution$Cu^{+} + e^- \to Cu$ ; $E^o = X_1 \ V$
$Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2 \ V$
Then for $Cu^{2+} + e^- \to Cu^{+}$ ; $E^o$ will be ?
$Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2 \ V$
Then for $Cu^{2+} + e^- \to Cu^{+}$ ; $E^o$ will be ?
Medium
View SolutionThe standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$
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