The E^{θ} of {M}_{{M}}²+} \| {Cu}²+} {Cu} is 0.3\ {V}. At what concentration of {Cu}²+} (. in {mol}\ {L}^{-1}), the E_{\;{cell }\;} value becomes zero? ≤ft(2.303\,RT/F=0.06)\;;\; (Conc. of {M}²+}=0.1\ {M} ) [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 10−11\;10^{-11}10−11

Correct Answer is : 10−11\;10^{-11}10−11

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Electrochemistry

Section: Chemistry

Question : 20 of 20

Marks: +1, -0

The

E^{\theta}

\mathrm{M}_{\mathrm{M}}^{2+} \| \mathrm{Cu}^{2+} \mid \mathrm{Cu}

0.3\ \mathrm{V}

\mathrm{Cu}^{2+} (.

\mathrm{mol}\ \mathrm{L}^{-1})

E_{\;\text{cell }\;}

\left(\frac{2.303\,RT}{F}=0.06\right)\;;\;

\mathrm{M}^{2+}=0.1\ \mathrm{M}

Solution:

The given standard electrode potential is:

E_{\;\text{cell }\;}^{\circ}=0.3\ \mathrm{V}

According to the Nernst equation:

E_{\;\text{cell }\;} = E_{\;\text{cell }\;}^{\circ} - \frac{2.303\,R T}{n F} \log \left[ \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{M}^{2+}]} \right]

In this context:

n=2

(since it involves the transfer of 2 electrons)

E_{\;\text{cell }\;}=0\;\text{ (when the cell voltage is zero)}\;

\;[\mathrm{M}^{2+}] =0.1\ \mathrm{M}

\;\frac{2.303\,R T}{F}=0.06

Substituting these values into equation (1) gives:

0.3 = - \frac{0.06}{2} \log \left[ \frac{[\mathrm{Cu}^{2+}]}{0.1} \right]

Solving for the log term:

-\frac{0.6}{0.06} = \log \left[ \frac{[\mathrm{Cu}^{2+}]}{0.1} \right]

Taking the antilog:

10^{-10} = \frac{[\mathrm{Cu}^{2+}]}{0.1}

This implies:

[\mathrm{Cu}^{2+}] = 10^{-10} \times 10^{-1} = 10^{-11} \text{mol/L}

Therefore, the concentration of

\mathrm{Cu}^{2+}

required to make the cell potential zero is

10^{-11} \text{mol/L}

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