JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Medium

1 of 1

Application +4 | -1

The Gibbs energy change (in KJ) for the given reaction at

\left[ {C{u^{2 + }}} \right] = \left[ {S{n^{2 + }}} \right] = 1M

and 298 K is

Cu\left( s \right) + S{n^{2 + }}\left( {aq} \right) \to C{u^{2 + }}\left( {aq} \right) + SN\left( s \right);\,\left( {E_{S{n^{2 + }}/Sn}^0 = - 0.16V,\,E_{C{u^{2 + }}/Cu}^0 = 0.34V} \right)

Take,

F = 96500\,\,\,C\,\,mo{l^{ - 1}}

)

View answer and explanation

Correct option: A

96.5 KJ

Option A (96.5 KJ) is correct because it is the Gibbs free energy change calculated using the standard reduction potentials and Faraday's constant.

More Questions from ELECTRO CHEMISTRY

Practice All Chapter MCQs →

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