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  • <img alt="\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{2+}(0.001 \mathrm{M}) \rightarrow \mathrm{Cu}^{2+}(0.01 \mathrm{M})+\mathrm{Sn}(\mathrm{s})" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cm

\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{2+}(0.001 \mathrm{M}) \rightarrow \mathrm{Cu}^{2+}(0.01 \mathrm{M})+\mathrm{Sn}(\mathrm{s})

The Gibbs free energy change for the above reaction at 298 \mathrm{~K} \text { is } x \times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1} \text {. }

The value of x is _________.[nearest integer]

Given: \left.\mathrm{E}_{\mathrm{Cu}^{2}+/ \mathrm{Cu}}^{\ominus}=0.34 \mathrm{~V} ; \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\ominus}=-0.14 \mathrm{~V} ; \mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]

Option: 1

983


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\text{E}^{\text{o}}\text { for the given reection }=-0.14-0.34=-0.48

Applying Nernst equation.

\mathrm{E=E^{0}-\frac{0.059}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[Sn ^{2+}\right]}}

     \mathrm{=-0.48-0.0295\; \log 10}

      =-0.5095

\begin{aligned} \therefore \Delta \text{G=-n F E} &=2 \times 96500 \times 0.5095 \mathrm{~J} \\&=98335 \mathrm{~J} \\&=98.34 \mathrm{~kJ} \end{aligned}

Hence, the answer is 983.

Posted by

shivangi.bhatnagar

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