Question
The standard reduction potential for the half-cell reaction, $$C{l_2} + 2{e^ - } \to 2C{l^ - }$$ will be $$\left( {P{t^{2 + }} + 2C{l^ - } \to Pt + C{l_2},E_{{\text{cell}}}^ \circ = - 0.15\,V;P{t^{2 + }} + 2{e^ - } \to Pt,{E^ \circ } = 1.20\,V} \right)$$
A.
- 1.35 $$V$$
B.
+ 1.35 $$V$$
C.
- 1.05 $$V$$
D.
+ 1.05 $$V$$
Answer :
+ 1.35 $$V$$
Solution :
$$\eqalign{
& Pt + C{l_2} \to P{t^{2 + }} + 2C{l^ - };E_{{\text{cell}}}^ \circ = 0.15\,V \cr
& \underline {P{t^{2 + }} + 2{e^ - } \to Pt;{E^ \circ } = 1.20\,V\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr
& \underline {C{l_2} + 2{e^ - } \to 2C{l^ - };{E^ \circ } = 1.35\,V\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr} $$