A) \[{{10}^{41}}\]
B) \[{{10}^{32}}\]
C) \[{{10}^{-32}}\]
D) \[{{10}^{-42}}\]
Correct Answer: B
Solution :
[b] \[S{{n}^{4+}}+2{{e}^{-}}\xrightarrow{{}}S{{n}^{2+}}E{}^\circ =0.13\,V\] \[B{{r}_{2}}+2{{e}^{-}}\xrightarrow{{}}2B{{r}^{-}}E{}^\circ =1.08\,V\] \[E{}^\circ \]values shows\[B{{r}_{2}}\] has higher reduction potential. Hence \[{{E}_{cell}}={{E}_{R}}-{{E}_{L}}\] \[={{E}_{B{{r}_{2}}/B{{r}^{-}}}}-{{E}_{S{{n}^{+4}}/S{{n}^{+2}}}}\] \[=1.08-0.13=0.95\,V\] Now \[-\Delta G=nF{{E}_{cell}}\] \[n=2,\]\[F=96500\] \[-\Delta G=2\times 96500\times 0.95\,kJ/mol\] Also, \[\Delta G=-2.303\,RT\,\log \,\,{{K}_{eq}}\] \[\log \,\,{{K}_{eq}}=-\frac{\Delta G}{2.303\times R\times T}\] \[=\frac{-(-2\times 96500\times 0.95)}{2.303\times 8.314\times 293}=32.6820\] \[{{K}_{eq}}=\text{antilog}\,\text{32}\text{.682}\] \[=4.78\times {{10}^{32}}\approx {{10}^{32}}\]
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