MCQ
For the cell reaction, $2C{{e}^{4+}}+Co\to 2C{{e}^{3+}}+C{{o}^{2+}}$ $E_{cell}^o$ is $1.89\,V$ and $E_{Co/C{o^{2 + }}}^o = + 0.28.$ If $E_{C{e^{4 + }}/C{o^{3 + }}}^o$ ............. $\mathrm{V}$
- A$-1.61$
- ✓$+1.61$
- C$-2.08$
- D$+2.17$
✓
Answer
Correct option: B.
$+1.61$
b
$2 C e^{4+}+C o \rightarrow 2 C e^{3+}+C o^{2+}$
$2 C e^{4+}+C o \rightarrow 2 C e^{3+}+C o^{2+}$
Anode : $C o \rightarrow C o^{2+}+2 e^{-}$
Cathode : $2 C e^{4+}+2 e^{-} \rightarrow C e^{3+}$
$E_{c e l l}=E_{R}-E_{L}$
$E_{c e l l}=E_{C e^{4+} / C e^{3+}}-E_{C o^{2+} / C o}$
$1.89=E_{C e^{4+} / C e^{3+}}-(-0.28)$
$E_{C e^{4}+} / C e^{3+}=1.89-0.28$
$E_{C e^{4}+} / C e^{3+}=1.61 \,V$
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