Given E_{Fe^{3+}/Fe^{2+}}^{circ} = +0.76 V a | vedclass

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Question

(A) The cell reaction is $2Fe^{3+} + 2I^{-} \longrightarrow 2Fe^{2+} + I_{2}$.Here,the number of electrons transferred $n = 2$.The standard cell poten

Vedclass · Accepted Answer

$1 	imes 10^{7}$

Vedclass · Answer

(A) The cell reaction is $2Fe^{3+} + 2I^{-} \longrightarrow 2Fe^{2+} + I_{2}$.Here,the number of electrons transferred $n = 2$.The standard cell potential is $E_{	ext{cell}}^{\circ} = E_{	ext{cathode}}^{\circ} - E_{	ext{anode}}^{\circ} = 0.76 \ V - 0.55 \ V = 0.21 \ V$.The relationship between equilibrium constant $K_{c}$ and $E_{	ext{cell}}^{\circ}$ is given by $\log K_{c} = \frac{n E_{	ext{cell}}^{\circ}}{0.0591} \approx \frac{n E_{	ext{cell}}^{\circ}}{0.06}$.Substituting the values: $\log K_{c} = \frac{2 	imes 0.21}{0.06} = \frac{0.42}{0.06} = 7$.Therefore,$K_{c} = 10^{7} = 1 	imes 10^{7}$.

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+}\|Cu$	$(II)$ $0.059$
$(C)$ $Zn\|Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

Given $E_{Fe^{3+}/Fe^{2+}}^{\circ} = +0.76 \ V$ and $E_{I_{2}/I^{-}}^{\circ} = +0.55 \ V$. The equilibrium constant for the reaction taking place in a galvanic cell consisting of the above two electrodes is (Given $\frac{2.303 \ RT}{F} = 0.06 \ V$)

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