MCQ
For the cell reaction, $2C{e^{4 + }} + Co \to 2C{e^{3 + }} + C{o^{2 + }}$ $E{^\circ _{cell}}$ is $1.89\,V$. If $E{^\circ _{C{e^{4 + }}/C{e^{3 + }}}}$ ........... $\mathrm{V}$
- A$-1.64$
- ✓$+ 1.64$
- C$-2.08 $
- D$+ 2.17$
✓
Answer
Correct option: B.
$+ 1.64$
b
(b)In this cell $Co$ is oxidised and it acts as anode and $Ce$ acts as cathode.
(b)In this cell $Co$ is oxidised and it acts as anode and $Ce$ acts as cathode.
$E_{Cell}^0 = E_{Cathode}^0 - E_{Anode}^0$ $ = 1.89 = E_{Cell}^0 - ( - 0.28)$
$E_{Cell}^0 = 1.89 - 0.28 = 1.61$ Volts.
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