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  • The cell potential for the given cell at  <img alt="\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{b

The cell potential for the given cell at 298 \mathrm{~K}

\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{bar})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Cu}^{2+}(\mathrm{aq}) \mid \mathrm{Cu}(\mathrm{s})

is 0.31 \mathrm{~V} \text {. } The pH of the acidic solution is found to be 3, whereas the concentration of \mathrm{Cu}^{2+} is 10^{-x} \mathrm{M}. The value of x is ____.

\text { (Given: } \left.\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.34 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The reaction occuring in the given cell is 

\mathrm{\mathrm{H}_{2}(g)+\mathrm{Cu}^{2+} \text { (aq.) } \longrightarrow 2H^{+} \text {(aq.) }+\mathrm{Cu}(s)}

Applying the Nernst equation, we have 

\mathrm{0.31=0.34-\frac{0.06}{2} \log \frac{\left[\mathrm{H}^{+}\right]^{2}}{\left[\mathrm{Cu}^{2+}\right]}}

\mathrm{0.31=0.34-\frac{0.06}{2} \log \frac{\left[10^{-3}\right]^{2}}{\left[Cu^{2+}\right]}}

\mathrm{ {\left[Cu^{2 +}\right] } =10^{-7} \\ }\\

\mathrm{x =7}

Hence the answer is 7

Posted by

Anam Khan

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