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What will be the emf for the given cell

$Pt\left | H_{2} \left ( P_{1} \right )\right |H^{+}{_{aq}}\left \parallel H_{2} \left ( P_{2}\right )\mid Pt$

Option 1)
$\frac{RT}{F}\log \frac{P_{1}}{P_{2}}$

Option 2)
$\frac{RT}{2F}\log \frac{P_{1}}{P_{2}}$

Option 3)
$\frac{RT}{F}\log \frac{P_{2}}{P_{1}}$

Option 4)
none of these.

Answers (1)

best_answer

As we learnt in

Half Reaction Method -

1. Produce unbalanced equation for the reaction in ionic form.

2. Separate the equation into half reactions.

3. Balance the atoms other than O and H in each half reaction.

4. Add $H_{2}O$ to balance O atoms and $H^{+}$ to balance H atom.

5. Add electrons to one side of the half reaction to balance the charges.

-

Electrode Potential(Nerst Equation) -

$M^{n+}(aq)+ne^{-}\rightarrow M(s)$

- wherein

$E_{(M^{n+}/M)}=E_{(M^{n+}/M)}^{0}-\frac{RT}{nf}ln\frac{[M]}{[M^{n+}]}$

$[M^{n+}]$ is concentration of species

F= 96487 C $moi^{-1}$

R= 8.314 $JK^{-1}moi^{-1}$

T= Temperature in kelvin

Lets break the reaction into half alls.

$2H^{+}+2c^{-}\rightarrow H_{2}(P_{1}), E_{1}$

where $E_{1}$ is potential at anode
also $2H^{+}+2c^{-}\rightarrow H_{2}(P_{2}), E_{2}$

where $E_{2}$ is potential at cathode

according to the Nernst equation, we have

$\\ E_{1}= E_{^{\circ}}\frac{ - RT}{2F} ln \left [ \frac{P_{1}}{c} \right ]\\ E_{1}= E_{^{\circ}}\frac{ - RT}{2F} ln \left [ \frac{P_{2}}{c} \right ]$

here $[ H^{+} ] = C$

emf of all, $E= E_{2} - E_{1}$

$= \frac{-RT}{2F} log \left [ \frac{P_{2}}{P_{1}} \right ]= \frac{RT}{2F} log \left [ \frac{P_{1}}{_{P2}} \right ]$

Correct option is 2.

Option 1)

$\frac{RT}{F}\log \frac{P_{1}}{P_{2}}$

Incorrect

Option 2)

$\frac{RT}{2F}\log \frac{P_{1}}{P_{2}}$

Correct

Option 3)

$\frac{RT}{F}\log \frac{P_{2}}{P_{1}}$

Incorrect

Option 4)

none of these.

Incorrect

Posted by

satyajeet

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