MCQ
The $IUPAC$ name of given is :
- ✓cis-cis$-9, 12-$ octadecandienoic acid
- Bcis-trans$-9,12-$ octadecan dienoic acid
- C$9,10-$ octa decadienoic acid
- D$9,14-$ octa decadienoic acid
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${\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}, \quad\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, \quad\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},}$
${\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}}$
[Given atomic numbers: $\mathrm{V}=23, \mathrm{Cr}=24, \mathrm{Fe}=26$,
$\mathrm{Ni}=28, \mathrm{Cu}=29]$
$Mn_{(S)} | Mn^{+2}_{(aq)} (0.4\,M) | | Sn^{+2}_{(aq)} (0.04\,M)| Sn_{(S)}$,
Calculate free energy change $\left( {\Delta G} \right)$ at $298\, K.$ .......... $\mathrm{kJ}$
Given : $E_{M{n^{ + 2}}/Mn}^o = - 1.18\,V ; \,E_{S{n^{ + 2}}/Sn}^o = - 0.14\,{\rm{volt}}\,\frac{{2.303RT}}{F}=0.06$