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  • <img alt="\operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_2(\mathrm{~g})(1 \mathrm{bar})\left\|\mathrm{H}^{+}(\mathrm{aq})(1 \mathrm{M})\right\| \mathrm{M}^{3+}(\mathrm{aq}), \mathrm{M}^{+}(\mathrm{a

\operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_2(\mathrm{~g})(1 \mathrm{bar})\left\|\mathrm{H}^{+}(\mathrm{aq})(1 \mathrm{M})\right\| \mathrm{M}^{3+}(\mathrm{aq}), \mathrm{M}^{+}(\mathrm{aq})\right| \operatorname{Pt}(\mathrm{s})
The \mathrm{E}_{\text {cell }} for the given cell is 0.1115 \mathrm{~V}at 298 \mathrm{~K} when \frac{\left[M^{+}(a q)\right]}{\left[M^{3+}(a q)\right]}=10^a 

The value of a is

\text { Given : } \mathrm{E}^\theta \mathrm{M}^{3+} / \mathrm{M}^{+}=0.2 \mathrm{~V}
\frac{2.303 R T}{F}=0.059 \mathrm{~V}

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Cell Reaction

\mathrm{H}_2+\mathrm{M}^{3+} \longrightarrow 2 \mathrm{H}^{+}+\mathrm{M}^{+}

\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{2.303 \mathrm{RT}}{2 \mathrm{~F}} \log \frac{\left[\mathrm{M}^{+}\right]\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{M}^{3+}\right]}
0.1115=0.2-\frac{0.059}{2} \log 10^{\mathrm{a}}
\frac{0.059}{2} \log 10^{\mathrm{a}}=0.0885
\mathrm{a}=3

Posted by

Ritika Kankaria

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