JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Easy

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Application +4 | -1

Consider the following cell representation:

\mathrm{Pt}(\mathrm{s})\left|\mathrm{Br}_2({l})\right| \mathrm{Br}^{\ominus}(0.010)|| \mathrm{H}^{\oplus}(0.030 \mathrm{~M})\left|\mathrm{H}_2(\mathrm{g})(1 \mathrm{~bar})\right| \mathrm{Pt}(\mathrm{s})

The EMF of the above cell at

298 \mathrm{~K}

-\mathrm{xV}

. Find

3 \mathrm{x}

(nearest integer)

\begin{aligned} & \text {[Given } \mathrm{E}^{\circ}_{\frac{1}{2}\mathrm{Br}_2 \mid \mathrm{Br}^{-}} \text{ (Reduction potential of } \mathrm{Br}_2)=+1.08 \mathrm{~V} \text {]} \end{aligned}

View answer and explanation

Correct option: A

Option A (4) is correct because the EMF of the cell is calculated using the Nernst equation, and then multiplied by 3 as requested.

More Questions from ELECTRO CHEMISTRY

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25^{\circ} \mathrm{C}

0.295 \mathrm{~V}

{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

\mathrm{CuS}{\mathrm{O}_4}

{AlCl}_3

STP

({F}=96500 {C})

2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Ag}_2 \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-}

\mathrm{pH}=3

0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S},

\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}}

\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48},

E^{\circ}_{{\mathrm{Ag} | \mathrm{Ag}}^{+}}=0.80 \mathrm{~V},

\log 2=0.3,

2.303 R T / F=0.06