Consider the following cell reaction:2 Fe_{(s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The Nernst equation for the given cell reaction is:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$Here,$n = 4$ (number of electrons transf

Vedclass · Accepted Answer

$1.57$

Vedclass · Answer

(D) The Nernst equation for the given cell reaction is:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$Here,$n = 4$ (number of electrons transferred).The reaction quotient $Q$ is given by:$Q = \frac{[Fe^{2+}]^2}{P(O_2) 	imes [H^+]^4}$Given $[Fe^{2+}] = 10^{-3} \ M$,$P(O_2) = 0.1 \ atm$,and $pH = 3$,so $[H^+] = 10^{-3} \ M$.$Q = \frac{(10^{-3})^2}{0.1 	imes (10^{-3})^4} = \frac{10^{-6}}{0.1 	imes 10^{-12}} = \frac{10^{-6}}{10^{-13}} = 10^7$Now,substitute the values into the Nernst equation:$E_{cell} = 1.67 - \frac{0.0591}{4} \log(10^7)$$E_{cell} = 1.67 - \frac{0.0591 	imes 7}{4}$$E_{cell} = 1.67 - 0.1034 \approx 1.57 \ V$

Consider the following cell reaction:
$2 Fe_{(s)} + O_{2_{(g)}} + 4 H_{(aq)}^{+} \rightarrow 2 Fe_{(aq)}^{2+} + 2 H_2O_{(l)} \quad E^{\circ} = 1.67 \ V$. At $[Fe^{2+}] = 10^{-3} \ M, P(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25^{\circ} C$ is (in $V$)

Similar Questions

If the $E^{\circ}_{cell}$ for a given reaction has a negative value,which of the following gives the correct relationships for the values of $\Delta G^{\circ}$ and $K_{eq}$ ?

For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

For a given half-cell,$Al^{3+} + 3e^{-} \rightarrow Al$,on increasing the concentration of aluminium ions,the electrode potential will

What will be the oxidation potential for the following hydrogen half cell at $1 \ bar$ pressure and $25^{\circ} C$ temperature (in $V$)?
$Pt \mid H_{2(g)} (1 \ bar) \mid HCl_{(aq)} \ pH = 3$

Determine the electrode potential of $Sn^{2+}(0.01 \text{ M}) | Sn(s)$ at $25^{\circ} \text{C}$ if $E^0_{Sn^{2+}/Sn}$ is $-0.136 \text{ V}$. (in $\text{ V}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider the following cell reaction:$2 Fe_{(s)} + O_{2_{(g)}} + 4 H_{(aq)}^{+} \rightarrow 2 Fe_{(aq)}^{2+} + 2 H_2O_{(l)} \quad E^{\circ} = 1.67 \ V$. At $[Fe^{2+}] = 10^{-3} \ M, P(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25^{\circ} C$ is (in $V$)