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.

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