A) \[{{10}^{41}}\]
B) \[{{10}^{32}}\]
C) \[{{10}^{-32}}\]
D) \[{{10}^{-42}}\]
Correct Answer: B
Solution :
| \[{{\operatorname{Sn}}^{4}}+2{{e}^{-}}\xrightarrow{{}}S{{n}^{2+}}:E{}^\circ =0.13V\]\[{{\operatorname{Br}}_{2}}+2{{e}^{-}}\xrightarrow{{}}2B{{r}^{-}};E{}^\circ =1.08V\] |
| \[\operatorname{E}{}^\circ \]Value shown \[{{\operatorname{Br}}_{2}}\]has higher reduction potential. |
| Hence, |
| \[{{\operatorname{E}}_{cell}}={{E}_{B{{r}_{2}}/B{{r}^{-}}}}-{{E}_{s{{n}^{+4}}/s{{n}^{+2}}}}\] |
| \[=1.08-0.13=0.95V\] |
| Now \[-\Delta \operatorname{G}=nF{{E}_{cell}}\] |
| \[n=2,\operatorname{F}=96500\] |
| \[-\Delta \operatorname{G}=2\times 96500\times 0.95kJ/mol.\] |
| Also, \[\Delta \operatorname{G}=-2.303RT\,log\,{{K}_{\operatorname{eq}}}\] |
| log \[{{\operatorname{K}}_{\operatorname{eq}}}=antilog32.682\approx {{10}^{32}}\] |
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