Consider the following cell reaction:2 Fe^{3+ | vedclass

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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 be

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The cell reaction is $2 Fe^{3+} + 2 I^{-} ightleftharpoons 2 Fe^{2+} + I_2$.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 $n = 2$:$0.237 = \frac{0.0591}{2} \log K_{eq}$.$\log K_{eq} = \frac{0.237 	imes 2}{0.0591} \approx 8.02$.Since $K_{eq} = 10^x$,we have $\log K_{eq} = x$.Therefore,$x \approx 8$.

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

For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

In the following reaction,what is the value of equilibrium constant?
$Cu_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+} + 2 Ag_{(s)}$
$E_{cell}^0 = 0.46 \ V$

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

Calculate the cell potential at $298 \ K$ for the following cell:
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)

Which of the following equations represents the correct relationship between the standard cell potential and the equilibrium constant for a cell reaction?

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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:$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$