MCQ
જ્યારે $C{H_2} = CH - COOH$ is reduced with $LiAl{H_4}$, the compound obtained will be
- A$C{H_3} - C{H_2} - COOH$
- ✓$C{H_2} = CH - C{H_2}OH$
- C$C{H_3} - C{H_2} - C{H_2}OH$
- D$C{H_3} - C{H_2} - CHO$
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$(i)\, Cu^{2+} + 2e^- \rightarrow Cu\,,$ $ E^o = 0.337\, V$
$(ii)\, Cu^{2+} + e^- \rightarrow Cu^+\,,$ $ E^o = 0.153\, V$
Electrode potential, $E^o$ for the reaction,
$Cu^+ + e^- \rightarrow Cu\,,$ will be ............ $V$.