Using the standard electrode potentials given in the Table $8.1$,predict if the reaction between the following is feasible:
$(a)$ $Fe^{3+}_{(aq)}$ and $I^{-}_{(aq)}$
$(b)$ $Ag^{+}_{(aq)}$ and $Cu_{(s)}$
$(c)$ $Fe^{3+}_{(aq)}$ and $Cu_{(s)}$
$(d)$ $Ag_{(s)}$ and $Fe^{3+}_{(aq)}$
$(e)$ $Br_{2(aq)}$ and $Fe^{2+}_{(aq)}$
$(a)$ $Fe^{3+}_{(aq)}$ and $I^{-}_{(aq)}$
$(b)$ $Ag^{+}_{(aq)}$ and $Cu_{(s)}$
$(c)$ $Fe^{3+}_{(aq)}$ and $Cu_{(s)}$
$(d)$ $Ag_{(s)}$ and $Fe^{3+}_{(aq)}$
$(e)$ $Br_{2(aq)}$ and $Fe^{2+}_{(aq)}$
(N/A) The possible reaction between $Fe^{3+}_{(aq)} + I^{-}_{(aq)}$ is $2Fe^{3+}_{(aq)} + 2I^{-}_{(aq)} \to 2Fe^{2+}_{(aq)} + I_{2(s)}$.
Oxidation: $2I^{-}_{(aq)} \to I_{2(s)} + 2e^{-}; E^{\circ} = -0.54 \ V$
Reduction: $[Fe^{3+}_{(aq)} + e^{-} \to Fe^{2+}_{(aq)}] \times 2; E^{\circ} = +0.77 \ V$
Overall: $E^{\circ}_{cell} = +0.77 - 0.54 = +0.23 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(b)$ The possible reaction between $Ag^{+}_{(aq)} + Cu_{(s)}$ is $2Ag^{+}_{(aq)} + Cu_{(s)} \to 2Ag_{(s)} + Cu^{2+}_{(aq)}$.
Oxidation: $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^{-}; E^{\circ} = -0.34 \ V$
Reduction: $[Ag^{+}_{(aq)} + e^{-} \to Ag_{(s)}] \times 2; E^{\circ} = +0.80 \ V$
Overall: $E^{\circ}_{cell} = +0.80 - 0.34 = +0.46 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(c)$ The possible reaction between $Fe^{3+}_{(aq)} + Cu_{(s)}$ is $2Fe^{3+}_{(aq)} + Cu_{(s)} \to 2Fe^{2+}_{(aq)} + Cu^{2+}_{(aq)}$.
Oxidation: $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^{-}; E^{\circ} = -0.34 \ V$
Reduction: $[Fe^{3+}_{(aq)} + e^{-} \to Fe^{2+}_{(aq)}] \times 2; E^{\circ} = +0.77 \ V$
Overall: $E^{\circ}_{cell} = +0.77 - 0.34 = +0.43 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(d)$ The possible reaction between $Ag_{(s)} + Fe^{3+}_{(aq)}$ is $Ag_{(s)} + Fe^{3+}_{(aq)} \to Ag^{+}_{(aq)} + Fe^{2+}_{(aq)}$.
Oxidation: $Ag_{(s)} \to Ag^{+}_{(aq)} + e^{-}; E^{\circ} = -0.80 \ V$
Reduction: $Fe^{3+}_{(aq)} + e^{-} \to Fe^{2+}_{(aq)}; E^{\circ} = +0.77 \ V$
Overall: $E^{\circ}_{cell} = +0.77 - 0.80 = -0.03 \ V$. Since $E^{\circ}_{cell} < 0$,the reaction is not feasible.
$(e)$ The possible reaction between $Br_{2(aq)} + Fe^{2+}_{(aq)}$ is $Br_{2(aq)} + 2Fe^{2+}_{(aq)} \to 2Br^{-}_{(aq)} + 2Fe^{3+}_{(aq)}$.
Oxidation: $[Fe^{2+}_{(aq)} \to Fe^{3+}_{(aq)} + e^{-}] \times 2; E^{\circ} = -0.77 \ V$
Reduction: $Br_{2(aq)} + 2e^{-} \to 2Br^{-}_{(aq)}; E^{\circ} = +1.09 \ V$
Overall: $E^{\circ}_{cell} = +1.09 - 0.77 = +0.32 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
Oxidation: $2I^{-}_{(aq)} \to I_{2(s)} + 2e^{-}; E^{\circ} = -0.54 \ V$
Reduction: $[Fe^{3+}_{(aq)} + e^{-} \to Fe^{2+}_{(aq)}] \times 2; E^{\circ} = +0.77 \ V$
Overall: $E^{\circ}_{cell} = +0.77 - 0.54 = +0.23 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(b)$ The possible reaction between $Ag^{+}_{(aq)} + Cu_{(s)}$ is $2Ag^{+}_{(aq)} + Cu_{(s)} \to 2Ag_{(s)} + Cu^{2+}_{(aq)}$.
Oxidation: $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^{-}; E^{\circ} = -0.34 \ V$
Reduction: $[Ag^{+}_{(aq)} + e^{-} \to Ag_{(s)}] \times 2; E^{\circ} = +0.80 \ V$
Overall: $E^{\circ}_{cell} = +0.80 - 0.34 = +0.46 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(c)$ The possible reaction between $Fe^{3+}_{(aq)} + Cu_{(s)}$ is $2Fe^{3+}_{(aq)} + Cu_{(s)} \to 2Fe^{2+}_{(aq)} + Cu^{2+}_{(aq)}$.
Oxidation: $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^{-}; E^{\circ} = -0.34 \ V$
Reduction: $[Fe^{3+}_{(aq)} + e^{-} \to Fe^{2+}_{(aq)}] \times 2; E^{\circ} = +0.77 \ V$
Overall: $E^{\circ}_{cell} = +0.77 - 0.34 = +0.43 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
$(d)$ The possible reaction between $Ag_{(s)} + Fe^{3+}_{(aq)}$ is $Ag_{(s)} + Fe^{3+}_{(aq)} \to Ag^{+}_{(aq)} + Fe^{2+}_{(aq)}$.
Oxidation: $Ag_{(s)} \to Ag^{+}_{(aq)} + e^{-}; E^{\circ} = -0.80 \ V$
Reduction: $Fe^{3+}_{(aq)} + e^{-} \to Fe^{2+}_{(aq)}; E^{\circ} = +0.77 \ V$
Overall: $E^{\circ}_{cell} = +0.77 - 0.80 = -0.03 \ V$. Since $E^{\circ}_{cell} < 0$,the reaction is not feasible.
$(e)$ The possible reaction between $Br_{2(aq)} + Fe^{2+}_{(aq)}$ is $Br_{2(aq)} + 2Fe^{2+}_{(aq)} \to 2Br^{-}_{(aq)} + 2Fe^{3+}_{(aq)}$.
Oxidation: $[Fe^{2+}_{(aq)} \to Fe^{3+}_{(aq)} + e^{-}] \times 2; E^{\circ} = -0.77 \ V$
Reduction: $Br_{2(aq)} + 2e^{-} \to 2Br^{-}_{(aq)}; E^{\circ} = +1.09 \ V$
Overall: $E^{\circ}_{cell} = +1.09 - 0.77 = +0.32 \ V$. Since $E^{\circ}_{cell} > 0$,the reaction is feasible.
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Similar Questions
Consider a $70 \%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1 \ bar$ and $298 \ K$. Its cell reaction is
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O(\ell)$
The work derived from the cell on the consumption of $1.0 \times 10^{-3} \ mol$ of $H_{2(g)}$ is used to compress $1.00 \ mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$) of the ideal gas?
The standard reduction potentials for the two half-cells are given below.
$O_{2(g)} + 4H^{+}(aq.) + 4e^- \rightarrow 2H_2O(\ell), E^{\circ} = 1.23 \ V$
$2H^{+}(aq.) + 2e^- \rightarrow H_{2(g)}, E^{\circ} = 0.00 \ V$
Use $F = 96500 \ C \ mol^{-1}, R = 8.314 \ J \ mol^{-1} \ K^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O(\ell)$
The work derived from the cell on the consumption of $1.0 \times 10^{-3} \ mol$ of $H_{2(g)}$ is used to compress $1.00 \ mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$) of the ideal gas?
The standard reduction potentials for the two half-cells are given below.
$O_{2(g)} + 4H^{+}(aq.) + 4e^- \rightarrow 2H_2O(\ell), E^{\circ} = 1.23 \ V$
$2H^{+}(aq.) + 2e^- \rightarrow H_{2(g)}, E^{\circ} = 0.00 \ V$
Use $F = 96500 \ C \ mol^{-1}, R = 8.314 \ J \ mol^{-1} \ K^{-1}$
MediumIIT 2020
View SolutionThe equation that is incorrect is
DifficultJEE MAIN 2020
View SolutionCalculate the equilibrium constant at $298 \ K$ for the following reaction and also calculate the maximum work that can be obtained from this cell:
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Given: $E_{Mg^{2+} \mid Mg}^{o} = -2.37 \ V$ and $E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V$
$Mg(s) \ | \ Mg^{2+}(aq) \ || \ Ag^{+}(aq) \ | \ Ag(s)$
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Consider the following data.
Electrolyte $\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$ $BaCl_2$ $x_1$ $H_2SO_4$ $x_2$ $HCl$ $x_3$
$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)
| Electrolyte | $\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$ |
|---|---|
| $BaCl_2$ | $x_1$ |
| $H_2SO_4$ | $x_2$ |
| $HCl$ | $x_3$ |
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DifficultJEE MAIN 2026
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$A$. Leclanche cell $1$. Converts energy of combustion into electrical energy $B$. Fuel cell $2$. Rechargeable cell $C$. Ni-Cd cell $3$. At anode,$Zn \longrightarrow Zn^{2+} + 2e^{-}$
| $A$. Leclanche cell | $1$. Converts energy of combustion into electrical energy |
| $B$. Fuel cell | $2$. Rechargeable cell |
| $C$. Ni-Cd cell | $3$. At anode,$Zn \longrightarrow Zn^{2+} + 2e^{-}$ |
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