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The cell,

$Zn\mid Zn^{2+}\left ( 1M \right )\parallel Cu^{2+}\left ( 1M \right )\mid Cu\left ( E^{\circ}{_{cell}}= 1.10V \right )$

was allowed to be completely discharged at 298 K .The relative concentration of $Zn^{2+}\: to\: Cu^{2+}$

$\left ( \frac{\left [ Zn^{2+} \right ]}{\left [ Cu^{2+} \right ]} \right )$ is

Option 1)
$9.65\times 10^{4}$

Option 2)
$antilog\left ( 24.08 \right )$

Option 3)
$37.3$

Option 4)
$10^{37.3}$

Answers (1)

As we learnt in

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

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

$E_{cell}=E^{\circ}_{cell}-\frac{0.059}{2}log\frac{\left [ Zn^{2+} \right ]}{\left [ Cu^{2+} \right ]}$

When the cell is completely discharged, $E_{cell}=0$

$0=1.1-\frac{0.059}{2}log\frac{\left [ Zn^{2+} \right ]}{\left [ Cu^{2+} \right ]}$

$or\; \; \; log\frac{\left [ Zn^{2+} \right ]}{\left [ Cu^{2+} \right ]}=\frac{2\times 1.1}{0.059}\; or,\; log\frac{Zn^{2+}}{Cu^{2+}}=37.3$

$or\; \; \; \frac{Zn^{2+}}{Cu^{2+}}=10^{37.3}$

Option 1)

$9.65\times 10^{4}$

This option is incorrect.

Option 2)

$antilog\left ( 24.08 \right )$

This option is incorrect.

Option 3)

$37.3$

This option is incorrect.

Option 4)

$10^{37.3}$

This option is correct.

Posted by

Sabhrant Ambastha

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