(A) The cell reaction is $2 Fe^{3+} + 2 I^{-} \rightleftharpoons 2 Fe^{2+} + I_2$.Here,the number of electrons transferred,$n = 2$.The relationship be

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(A) The cell reaction is $2 Fe^{3+} + 2 I^{-} \rightleftharpoons 2 Fe^{2+} + I_2$.Here,the number of electrons transferred,$n = 2$.The relationship between the equilibrium constant $K_{eq}$ and the standard cell potential $E^{\circ}_{cell}$ is given by the Nernst equation at equilibrium:$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$.Given $E^{\circ}_{cell} = 0.237 \ V$ and $n = 2$:$0.237 = \frac{0.0591}{2} \log K_{eq}$.$\log K_{eq} = \frac{0.237 \times 2}{0.0591} \approx 8.02$.Since $K_{eq} = 10^x$,we have $\log K_{eq} = x$.Therefore,$x \approx 8$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

Consider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$

AP EAMCET 2025 All AP EAMCET

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$8$
B
$7$
C
$6$
D
$9$

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Nernst equation and ECS

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What will be the electromotive force of the following cell at $298 \ K$ ............. $V$
$Pt \, | \, Br_{2(\ell), 0.1M} \, | \, Br^{-}_{(aq), 0.1M} \, || \, H^{+}_{(aq), 0.1M} \, | \, H_{2(g), 1atm} \, | \, Pt$
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Medium

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Calculate the equilibrium constant of the following reaction:
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Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)

Medium

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If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
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Difficult

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If $E^{\circ}(Cu^{2+}_{(aq)} \mid Cu_{(s)}) = +0.34 \ V$. What is the potential for $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$ at $298 \ K$?

MediumMHT CET 2025

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Consider the following cell reaction:$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$

Similar Questions

What will be the electromotive force of the following cell at $298 \ K$ ............. $V$$Pt \, | \, Br_{2(\ell), 0.1M} \, | \, Br^{-}_{(aq), 0.1M} \, || \, H^{+}_{(aq), 0.1M} \, | \, H_{2(g), 1atm} \, | \, Pt$Given: $E^{0}_{Br^{-}/Br_{2}} = -1.06 \ V$

Calculate the equilibrium constant of the following reaction:$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$, given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$

If $E^{\circ}(Cu^{2+}_{(aq)} \mid Cu_{(s)}) = +0.34 \ V$. What is the potential for $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$ at $298 \ K$?

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Consider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$

What will be the electromotive force of the following cell at $298 \ K$ ............. $V$
$Pt \, | \, Br_{2(\ell), 0.1M} \, | \, Br^{-}_{(aq), 0.1M} \, || \, H^{+}_{(aq), 0.1M} \, | \, H_{2(g), 1atm} \, | \, Pt$
Given: $E^{0}_{Br^{-}/Br_{2}} = -1.06 \ V$

Calculate the equilibrium constant of the following reaction:
$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$, given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

If $E_{cell} = 0.118 \, V$ for the following reaction,calculate $[H^{+}]$ and $pH$ at $298 \, K$ temperature.
$Pt \mid H_2(1 \, bar) \mid H^{+} (10^{-6} \, M) \parallel H^{+} (x \, M) \mid H_2 (1 \, bar) \mid Pt$